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Stoichiometric Calculations

Slide 3 / 109 Table of Contents

· Stoichiometry Calculations with Moles

Click on the topic to go to that section

· Stoichiometry Calculations with Particles and Volume · Stoichiometry Calculations with Mass · Mixed Stoichiometry Problems · Limiting Reactants · Theoretical, Actual and Percent Yield · Calculating Excess Reactants

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Stoichiometry Calculations with Moles

Slide 5 / 109 Stoichiometry

The word stoichiometry is derived from two Greek words. "stoicheion" meaning element and "metron", meaning measure.

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NaN3 capsule

Stoichiometry in our daily lives

Airbags save thousands of lives every

• year. When a collision happens, the

following reaction occurs. 2NaN3(s) --> 2Na(s) + 3N2(g) In order to properly inflate, roughly 50 L of nitrogen gas must be produced. Engineers have determined, using stoichiometry, that about 96 grams of NaN3 are needed to react in each airbag capsule to produce enough nitrogen gas.

Slide 7 / 109 Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

2 H2 + O2 --# 2 H2 O

can be read as: 2 moles of H2 plus 1 mole of O2 yields 2 moles of H2O giving ratios of: 2 mol H

2

2 mol H

2

1 mol O

2

1 mol O2 2 mol H2O 2 mol H2O

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N2 + 3H2 ---> 2NH3

The above ratios can be used to determine the quantity of any reactant and products. For every 1 mol of N2, · you would need 3 mol of H2 to completely react with · you would produce 2 mol of NH3

Stoichiometric Calculations

1 mol N2 3 mol H2 1 mol N2 3 mol H2 2 mol NH3 2 mol NH3

A balanced chemical equation is needed to perform any stoichiometric calculations.

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Using this interpretation it's straightforward to answer questions about the relative number of moles of reactants and products. For instance, use the balanced equation below to determine the maximum number of moles of H2O that could be created from reacting 8 moles of H2. 2 H2 + O2 2 H2O

Stoichiometric Calculations with Moles Slide 10 / 109

2 H2 + O2 2 H2O

• 1. Use the equation to set

up a ratio of the substances

• f interest.
• 2. Set that equal to the ratio
• f the known to unknown

quantities of the same substances.

• 3. Solve for the unknown.

2 mol H2O 2 mol H2 n mol H2O 8 mol H2 2 mol H2O 2 mol H2 = 8(2) = (2) n mol H2O n = 8 mol H2O

Stoichiometry Calculations with Moles Slide 11 / 109 Stoichiometry Calculations with Moles

Another Example: Given the equation: How many moles of oxygen would be needed to react with 12 moles of H2? 2 H2 + O2 2 H2O

Slide 12 / 109 Stoichiometry Calculations with Moles

2 H2 + O2 - -# 2 H2O

• 1. Use the formula to set up

a ratio of the substances of interest.

• 2. Set that equal to the ratio
• f the known to unknown

quantities of the same substances.

• 3. Solve for the unknown by

cross multiplying. 1 mol O2 2 mol H2 n mol O2 12 mol H2 1 mol O2 2 mol H2 = 12(1) = (2)n O2 = 6 n O2

Slide 13 / 109 Stoichiometry Calculations 2 H2 + O2 -# 2 H2O

Given 12 moles of H2, how much O2 and H2O would be needed or produced? Using the ratios, you would need..... 1/2 as much O2 as H2 (ratio is 1 mol O2/2 mol H2) 12 mol H2 x 1 mol O2 = 6 mol O2 needed 2 mol H2 Using the ratios, you would produce..... An equal amount of H2O as H2 used (ratio is 2 mol H2O/2 mol H2) 12 mol H2 x 2 mol H2O = 12 mol H2O 2 mol H2

Slide 14 / 109 Real World Application

2Al + Fe2O3 --> 2Fe + Al2O3 + 859 kJ of energy How many moles of Al would be needed to produce 7.8 moles of Al2O3? 7.8 mol Al2O3 x 2 mol Al = 15.6 mol Al 1 mol Al2O3

move for answer

The thermite reaction (above) releases a lot of heat and is used in to weld railroad tracks together.

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1 What is the largest number of moles of Al

2O3

that could result from reacting 6 moles of O2?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s)

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1 What is the largest number of moles of Al

2O3

that could result from reacting 6 moles of O2?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s) [This object is a pull tab]

Answer 4

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2 How many moles of O2 would be required to create 12 moles of Al

2 O3 ?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s)

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2 How many moles of O2 would be required to create 12 moles of Al

2 O3 ?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s) [This object is a pull tab]

Answer 18

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3 How many moles of O2 would be required to completely react with 8 moles of Al?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s)

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3 How many moles of O2 would be required to completely react with 8 moles of Al?

4 Al (s) + 3 O

2 (g) - - # 2 Al2O3 (s) [This object is a pull tab]

Answer 6

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4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol of Fe in this reaction? 4 Fe (s) + 3 O2 (g)--> 2 Fe2O3 (s)

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4 When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 2.4 mol of Fe in this reaction? 4 Fe (s) + 3 O2 (g)--> 2 Fe2O3 (s)

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Answer 1.8

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5 How many moles of aluminium are needed to react completely with 1.2 mol FeO? 2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al

2O3 (s)

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5 How many moles of aluminium are needed to react completely with 1.2 mol FeO? 2 Al (s) + 3 FeO (s) --> 3 Fe (s) + Al

2O3 (s)

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Answer 0.8

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6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) --> Ca (s) + Cl

2 (g)

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6 How many moles of calcium metal are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) --> Ca (s) + Cl

2 (g)

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Answer 8

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7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) --> Ca (s) + Cl

2 (g)

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7 How many moles (total) of calcium metal and chlorine gas are produced from the decomposition of 8 mol of calcium chloride? CaCl2 (s) --> Ca (s) + Cl

2 (g)

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Answer 16

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8 How many moles of Ag are needed to react with 40 moles of HNO3? 3 Ag(s) + 4 HNO3(aq) --> 3 AgNO3(aq) + 2 H2O(l) + NO(g)

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8 How many moles of Ag are needed to react with 40 moles of HNO3? 3 Ag(s) + 4 HNO3(aq) --> 3 AgNO3(aq) + 2 H2O(l) + NO(g)

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Answer 30

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9 How many moles of AgNO3 could be produced from 40 moles of HNO3? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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9 How many moles of AgNO3 could be produced from 40 moles of HNO3? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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Answer 30

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10 How many moles of water would be produced when 0.4 moles of Ag react with an excess amount of HNO3? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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10 How many moles of water would be produced when 0.4 moles of Ag react with an excess amount of HNO3? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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Answer 0.26

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11 How many moles of NO were produced if 16 moles of water were made during the reaction? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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11 How many moles of NO were produced if 16 moles of water were made during the reaction? 3Ag(s) + 4HNO3(aq) --> 3AgNO3(aq) + 2 H2O(l) + NO(g)

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Answer 8

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12 How many moles of N2H4 are required to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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12 How many moles of N2H4 are required to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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Answer 38

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13 How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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13 How many moles of dinitrogen tetraoxide would be needed to produce 57 moles of nitrogen gas? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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Answer 19

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14 How many moles of water are produced if 57 moles

• f nitrogen gas are produced?

2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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14 How many moles of water are produced if 57 moles

• f nitrogen gas are produced?

2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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Answer 76

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15 How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N2O4? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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15 How many total moles of gas would be produced when 5 moles of nitrogen tetrahydride reacts with excess N2O4? 2 N2H4(l) + N2O4(l) ---> 3 N2(g) + 4 H2O(g)

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Answer 17.5

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Stoichiometry Calculations with Particles and Volume

Slide 31 / 109 Stoichiometry Calculations with Particles

The number of particles (atoms, molecules, formula units) is directly to proportional to the number of moles. Therefore... 2 H2 + O2 --> 2 H2 O can be read as: 2 molecules of H2 plus 1 molecule of O2 yields 2 molecules of H2 O. Note...while moles can be expressed as non-whole numbers, particles must be whole numbers. One cannot have 6.1 atoms, molecules, or formula units!

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16 What is the largest number of of Li3 N formula units that could result from reacting 6 N2 molecules? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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16 What is the largest number of of Li3 N formula units that could result from reacting 6 N2 molecules? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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Answer

1 molecule N2 6 N2 molecules x 2 Li3N formula units = 12 Formula units Li3N

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17 How many N2 molecules would be required to create 4 Li3 N formula units? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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17 How many N2 molecules would be required to create 4 Li3 N formula units? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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Answer 2

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18 How many Li atoms would be required to completely react with 3 N2 molecules? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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18 How many Li atoms would be required to completely react with 3 N2 molecules? 6 Li (s) + N2 (g) --# 2 Li3 N (s)

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Answer 18

Slide 35 / 109 Stoichiometry Calculations with Volumes

At a given temperature and pressure, the space a sample of a gas takes up (it's volume) is proportional to the number of moles of gas molecules

• present. Therefore...

2 H2 (g) + O2 (g)

• -> 2 H2 O(g)

can be read as: 2 volumes of H2 plus 1 volume of O2 yields 2 volumes of H2 O. Note: The volume of a material is only proportional to the number of moles when the substance is in the gas phase!

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19The equation below shows the decomposition of lead nitrate. How many liters of oxygen are produced when 12L of NO2 are formed? (STP) 2Pb(NO3 )2 (s) --> 2PbO (s) +4NO2 (g) + O2 (g)

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19The equation below shows the decomposition of lead nitrate. How many liters of oxygen are produced when 12L of NO2 are formed? (STP) 2Pb(NO3 )2 (s) --> 2PbO (s) +4NO2 (g) + O2 (g)

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Answer

12 L NO2 x 1 L O2 = 3 L of O2

4 L NO2

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20 What volume of methane is needed to completely react with 500 mL of O2 at STP? (Balance the equation first!!!) __ CH4 + ___ O2 --# ___ CO2 + ___ H2 O

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20 What volume of methane is needed to completely react with 500 mL of O2 at STP? (Balance the equation first!!!) __ CH4 + ___ O2 --# ___ CO2 + ___ H2 O

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Answer 250

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21 How many liters of H2 O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount

• f O2 (g)?

2 H2 (g) + O2 (g) --# 2 H2 O (g)

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21 How many liters of H2 O (g) will be created from reacting 8.0 L of H2 (g) with a sufficient amount

• f O2 (g)?

2 H2 (g) + O2 (g) --# 2 H2 O (g)

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Answer 8

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22 How many liters of NO2 (g) will be created from reacting 36 L of O2 (g) with a sufficient amount of NH3 (g)? 4 NH3 (g) + 7 O2 (g) --# 4 NO2 (g) + 6 H2 O (g)

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22 How many liters of NO2 (g) will be created from reacting 36 L of O2 (g) with a sufficient amount of NH3 (g)? 4 NH3 (g) + 7 O2 (g) --# 4 NO2 (g) + 6 H2 O (g)

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Answer 20.6

Slide 40 / 109 Stoichiometry with Particles and Volumes

It's common to be asked to report a value in a unit other than the one given. For example: Given the following reaction, how many L of nitrogen gas would be needed to produce 3 moles of ammonia? N2(g) + 3H2(g) --> 2NH3

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Given the following reaction, how many liters of nitrogen gas would be needed @STP to produce 3 moles of ammonia? N2(g) + 3H2(g) --> 2NH3 mol NH3 --> mol N2 --> L N2 3 mol NH3 x 1 mol N2 x 22.4 L = 33.6 L N2 2 mol NH3 1 mol

Stoichiometry with Particles and Volumes Slide 42 / 109

Example 2: How many moles of Cl2 gas would be needed to produce 3 x 1024 formula units of NaCl given the following reaction. 2Na(s) + Cl2(g) --> 2NaCl(s) formula units NaCl --> molecules Cl2 --> mol Cl2 3 x 1024 formula units NaCl x 1 molecule Cl2 x 1 mol Cl2 2 for. units NaCl 6.02 x 1023 molecules = 2.5 mol Cl2

Stoichiometry with Particles and Volumes Slide 43 / 109

23 How many sodium atoms would be needed to react with 33.6 L of chlorine gas at STP? 2Na(s) + Cl2(g) --> 2NaCl Note: Sodium is a solid and therefore cannot be expressed in L, so first convert the chlorine gas to moles.

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23 How many sodium atoms would be needed to react with 33.6 L of chlorine gas at STP? 2Na(s) + Cl2(g) --> 2NaCl Note: Sodium is a solid and therefore cannot be expressed in L, so first convert the chlorine gas to moles.

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Answer

Strategize!! L Cl2 --> n Cl2 --> n Na --> atoms Na

33.6 L Cl2 x 1 mol Cl2 x 2 mol Na x 6.02 x 1023 atoms Na 22.4 L 1 mol Cl2 mol Na

= 1.8 x 1024 atoms Na

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24 How many liters of oxygen gas would need to be combusted with excess hydrocarbon to produce 5.5 moles of water @STP?

2 C57H110O6(s) + 163 O2(g) --> 114 CO

2(g) + 110 H 2O(l)

Note: The water product is a liquid not a gas, so it can't be converted to L. Find moles of oxygen gas first!

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24 How many liters of oxygen gas would need to be combusted with excess hydrocarbon to produce 5.5 moles of water @STP?

2 C57H110O6(s) + 163 O2(g) --> 114 CO

2(g) + 110 H 2O(l)

Note: The water product is a liquid not a gas, so it can't be converted to L. Find moles of oxygen gas first!

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Answer

Strategize!! mol H2O --> mol O2 --> L O2 5.5 mol H2O x 163 mol O2 x 22.4 L = 183 L O2 110 mol H2O 1 mol

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25 How many moles of carbon dioxide gas would be produced @STP if 2.24 L of O2 gas react?

2 C57H110O6(s) + 163 O2(g) --> 114 CO

2(g) + 110 H 2O(l)

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25 How many moles of carbon dioxide gas would be produced @STP if 2.24 L of O2 gas react?

2 C57H110O6(s) + 163 O2(g) --> 114 CO

2(g) + 110 H 2O(l)

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Answer 0.070

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26 How many L of water vapor would be produced @STP when 3.0 x 1018 molecules of hydrogen gas react? O2(g) + 2H2(g) --> 2H2O(g)

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26 How many L of water vapor would be produced @STP when 3.0 x 1018 molecules of hydrogen gas react? O2(g) + 2H2(g) --> 2H2O(g)

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Answer 1.12 x 10

4

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27 If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? 2Mg(s) + O2(g) --> 2MgO(s)

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27 If 44 moles of magnesium react, how many molecules of oxygen gas would be needed? 2Mg(s) + O2(g) --> 2MgO(s)

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Answer 1.35 x 10

25

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Stoichiometry Calculations with Mass

Slide 49 / 109 Mass Relationships in Stoichiometry

Unlike the volume, particles, and moles of a material which are independent of the type of material present, the mass of a material is specific to each substance and therefore different.

O2 gas H2 gas

1 mole 1 mole 22.4 L 22.4 L 6.02 x 1023 molecules 6.02 x 1023 molecules 32 grams 2 grams

Slide 50 / 109 Stoichiometry

Depending on the units, there are many ways to interpret a balanced equation! 2H2 + O2 -- # 2H2 O

2 molecules H2 + 1 molecule O2 -- # 2 molecules H2O 2 mol H2 + 1 mol O2 -- # 2 mol H2O 2 volumes H

2 + 1 volume O 2 --> 2 volumes H 2O

4.0 g H2 + 32.0 g O2 -- # 36.0 g H2O

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Example: How many grams of hydrogen gas would need to react with 3.4 moles of oxygen gas? 2H2(g) + O2(g) --> 2H2O(g) mol O2 --> mol H2 --> g H2 3.4 mol O2 x 2 mol H2 x 2 g H2 = 13.6 g H2 1 mol O2 1 mol H2

Mass Relationships in Stoichiometry

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Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if a product) or used (if a reactant).

aA bB

Mass-Mass Calculations

Grams of substance A Grams of substance B Moles of substance A Moles of substance B Use molar mass of A Use molar mass of B Use coefficients

• f A and B

from balanced equation Given Find

Slide 53 / 109 Mass-Mass Calculations

Example: Calculate the mass of ammonia, NH3 , produced by the reaction of 5.4 g hydrogen gas with an excess of nitrogen.

N2 + 3H2 ---> 2NH3

Strategize!! g H2 --> mol H

2 --> mol NH 3 --> g NH 3

5.4 g H2 x 1 mol H2 x 2 mol NH3 x 17 g NH3 2 g H2 3 mol H2 1 mol NH3

= 30.6 g NH3

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28 What is the mass of sodium produced when 40 grams of sodium azide decompose?

2 NaN3 (s) --> 2 Na (s) + 3 N2 (g)

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28 What is the mass of sodium produced when 40 grams of sodium azide decompose?

2 NaN3 (s) --> 2 Na (s) + 3 N2 (g)

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Answer 14.1

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29 How many grams of Al

2 O3 will be created

from reacting 36 g of Al with a sufficient amount of O2 ?

4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)

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29 How many grams of Al

2 O3 will be created

from reacting 36 g of Al with a sufficient amount of O2 ?

4 Al (s) + 3 O2 (g) --> 2 Al2O3 (s)

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Answer 68

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30 How many grams of Mg must react in

• rder to

to create 84 g of MgO?

2 Mg (s) + O2 (g) --> 2 MgO (s)

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30 How many grams of Mg must react in

• rder to

to create 84 g of MgO?

2 Mg (s) + O2 (g) --> 2 MgO (s)

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Answer 50.4

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Mixed Stoichiometry Problems

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Generally speaking, it is easiest to convert to moles first. mol A mol B L of A g of A particles

• f A

g of B L of B particles

• f B

Mixed Stoichiometry Problems Slide 59 / 109 Mixed Stoichiometry Calculations

Every type of stoichiometry calculation may be solved by following this map.

(1) From left to right, we convert any "Given" substance to moles. (2) Next, using the mole ratio created with coefficients, one can calculate the moles of the "Wanted" quantity. (3) Finally, if necessary, moles can be converted to either particles, mass or volume.

Slide 60 / 109 Mixed Stoichiometry Calculations

(1) (2) (3) representative particles of G 1 mol G 6.02 x 1023 mass

• f G

1 mol G mass G 1 mol G 22.4 L G volume of G at STP mol G b mol W a mol G mol W representative particles of W 6.02 x 1023 1 mol W mass W 1 mol W mass

• f W

22.4 L W 1 mol W Volume of W at STP = = = = = = x x x x x x = x

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31 How many L of water vapor can be produced from the combustion of 1 gram of glucose @STP?

C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(g)

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31 How many L of water vapor can be produced from the combustion of 1 gram of glucose @STP?

C6H12O6(s) + 6 O2(g) --> 6 CO2(g) + 6 H2O(g)

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Answer 0.75

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32 What mass of CaO would be required to completely react with 42 grams of H2O at STP?

CaO (s) + H2O (l) --> Ca(OH)2 (s)

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32 What mass of CaO would be required to completely react with 42 grams of H2O at STP?

CaO (s) + H2O (l) --> Ca(OH)2 (s)

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Answer 131

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33 How many grams of iron can be extracted from 500 kg

• f iron ore? (Make sure you balance the equation first

and remember to convert your kg --> g)

Fe2O3 (s) + C (s) --> Fe (s) + CO2 (g)

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33 How many grams of iron can be extracted from 500 kg

• f iron ore? (Make sure you balance the equation first

and remember to convert your kg --> g)

Fe2O3 (s) + C (s) --> Fe (s) + CO2 (g)

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Answer 350,000

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34 How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen gas @STP?

N2(g) + 3H2(g) --> 2NH3(g) Slide 64 (Answer) / 109

34 How many grams of ammonia can be produced by reacting 10 moles of nitrogen gas with excess hydrogen gas @STP?

N2(g) + 3H2(g) --> 2NH3(g)

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Answer 3.40

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35 How many L of O2 gas @STP are required to produce 90 grams of aluminum oxide? 4Al(s) + 3O2(g) --> 2Al2O3(s)

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35 How many L of O2 gas @STP are required to produce 90 grams of aluminum oxide? 4Al(s) + 3O2(g) --> 2Al2O3(s)

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Answer 29.6

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36 How many grams of chlorine gas are needed to react with 1 mole of Sb @STP?

2 Sb + 3 Cl2 --> 2 SbCl3

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36 How many grams of chlorine gas are needed to react with 1 mole of Sb @STP?

2 Sb + 3 Cl2 --> 2 SbCl3

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Answer 106.5

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37 How many moles of aluminum oxide are produced when 3.580 kg of manganese dioxide are consumed?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

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37 How many moles of aluminum oxide are produced when 3.580 kg of manganese dioxide are consumed?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

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Answer 27.4

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38 How many moles of manganese dioxide will be needed to react with 6 x 1025 atoms of Al?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

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38 How many moles of manganese dioxide will be needed to react with 6 x 1025 atoms of Al?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

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Answer 75

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39 If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

Slide 69 (Answer) / 109

39 If 4.37 moles of Al are consumed, how many formula units of aluminum oxide would be produced?

3MnO2(s) + 4 Al(s) --> 2 Al

2O3(s) + 3Mn(s)

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Answer 1.3 x 10

24

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Here 2 C57 H110 O6 (s) + 163 O2 (g) --> 114 CO

2 (g) + 110 H 2 O(l)

At STP, what volume of carbon dioxide is produced when 50 grams of tristearin is burned by the camel?

Real World Application

tristearin The compound tristearin (C57H110O6) is a type of fat which camels store in their hump and is used to make chocolate. (yes, really)

Slide 71 / 109

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Limiting Reactants

Slide 72 / 109 Concept of the Limiting Reactant

In a chemical reaction, do all of the reactants turn into products? What happens when one of the reactants gets used up? If the following reaction starts with 10 moles of H2 and 20 moles

• f Cl2, which reactant will run out of first?

H2(g) + Cl2(g) --> 2HCl(g)

Slide 73 / 109 Concept of the Limiting Reactant

The 10 moles of H2 are used to produce 20 moles of HCl. When the H2 is all used up, no more HCl can be produced. Were all the moles of Cl2 used up? How many are left over? H2(g) + Cl2(g) --> 2HCl(g)

Slide 74 / 109 Concept of the Limiting Reactant

No more HCl can be produced once the H2 runs out, therefore, H2 is the Limiting Reactant (limits the amount of product). Since the reaction started with more Cl2 than H2, not all of the Cl2 is used up in the reaction. Cl2 is the Excess Reactant (there's an excess amount). H2(g) + Cl2(g) --> 2HCl(g) Left-over Cl2 not used in the reaction to make HCL

Slide 75 / 109 Limiting Reactants

The limiting reactant, or limiting reagent, is the reactant present in the smallest stoichiometric amount. This is not necessarily the one with the smallest mass. The limiting reactant is the reactant you’ll run out of first, and it is the one that determines the maximum amount of product that can be made.

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Limiting reagent problems are worded differently because the quantities of both reactants are given. 10 moles of H2 and 20 moles of Cl2 react to produce HCl. Which quantity is the limiting reagent? It is your job to figure out which reactant is limiting because that will determine the maximum amount of product you can get, also called the maximum yield. There are a variety of methods to determine which reactant is the limiting one.

Limiting Reactants Slide 77 / 109 Steps to Determine the Limiting Reactant

A series of steps can be used to determine the limiting reactant in any reaction: Step 1: Convert the given quantities into moles. These are your initial amounts of each reactant. Step 2: Divide each by its stoichiometrical coefficient from the balanced chemical equation. This factors in how much is needed in the reaction. Step 3: Whichever reagent has the smallest quantity must be the limiting reactant!

Slide 78 / 109 Determining the Limiting Reactant

Example: When 10 grams of hydrogen react with 3.4 moles of nitrogen gas to make ammonia, which substance would be the limiting reactant? N2(g) + 3H2(g) --> 2NH3(g) Step 1: Convert all values to moles. 10 g H2 x 1 mol H2 = 5 mol H2 2 g H2 Initial Amounts = 5 mol H2, 3.4 mol N2

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Step 2: Find the stoichiometrical equivalents of each reactant Initial: 3.4 mol 5 mol Divide by coefficient: 3.4/1 5/3 Available Amounts: 3.4 mol 1.66 mol Step 3: Since there is less hydrogen gas, it will be the limiting reactant!

N2(g) + 3H2(g) --> 2NH3(g)

Determining the Limiting Reactant Slide 80 / 109 Limiting Reactants

Example: If 10 moles of hydrogen gas react with 7 moles of oxygen gas @STP to make water, which is the limiting reactant? 2H2(g) + O2(g) --> 2H2O(g) Step 1: Both amounts are already in moles! Step 2: 10 mol H2/2 = 5 mol H2 available 7 mol O2/1 = 7 mol O2 available Step 3: H2 gas is the limiting reactant. O2 gas will be left over and is the excess reactant.

Slide 81 / 109 Limiting Reactants

Example: Given that 2H2O(g) + O2(g) -->2H2O2(l) If 36 grams of water react with 44.8 L of oxygen gas @STP, which substance is the limiting reactant? Step 1: 36 g H2O x 1 mol = 2 mol H2O 44.8 L O2 x 1 mol = 2 mol O2 Step 2: 2 mol H2O/2 = 1 mol H2O available 2 mol O2/1 = 2 mol O2 available Step 3: Since less water is available, it is the limiting reactant. Oxygen gas is the excess reactant. 24 L O2 18 g H2O

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40 In this example, the ____ is the limiting reagent.

A

Hydrogen

B

Oxygen

Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2

C

water

Slide 82 (Answer) / 109

40 In this example, the ____ is the limiting reagent.

A

Hydrogen

B

Oxygen

Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2

C

water

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Answer A

Slide 83 / 109

41 In this example _______ is the excess reagent.

A

Hydrogen

B

Oxygen

C

Water

Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2

Slide 83 (Answer) / 109

41 In this example _______ is the excess reagent.

A

Hydrogen

B

Oxygen

C

Water

Before reaction 10H2 and 7 O2 After reaction 10 H2O and 2O2

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Answer B

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42 When 33 L of nitrogen gas react with 12 grams of hydrogen gas to make ammonia @STP, the hydrogen gas will be the excess reactant.

True False

N2(g) + 3H2(g) --> 2NH3(g)

Slide 84 (Answer) / 109

42 When 33 L of nitrogen gas react with 12 grams of hydrogen gas to make ammonia @STP, the hydrogen gas will be the excess reactant.

True False

N2(g) + 3H2(g) --> 2NH3(g)

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Answer True

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43 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn2+ + H2(g)

Slide 85 (Answer) / 109

43 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn2+ + H2(g)

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Answer False

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44 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn2+ + H2(g)

Slide 86 (Answer) / 109

44 When 120 grams of zinc react with 2.1 moles of H+ ion, the zinc will limit the reaction. True False Zn(s) + 2H+(aq) --> Zn2+ + H2(g)

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Answer False

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Real World Application

In your car engine, octane is combusted with oxygen to produce carbon dioxide and water (the exhaust). The mix of octane to oxygen must be right on or the mix is too rich (too much

• ctane) or too lean (too little octane).

If 0.065 L of oxygen is being mixed with 0.0061 L of octane @ STP, calculate if the mixture is running lean

• r running rich?

0.065 L x 1 = 0.0029 mol O2 0.0061 L x 1 = 0.00027 mol octane 0.0029/25 = 0.000116 O2 av. 0.00027 mol/2 = 0.00014 mol octane av. Excess of octane so mixture is too RICH!

move for answer

2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)

Slide 88 / 109

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Theoretical, Actual and Percent Yield

Slide 89 / 109 3 Types of Yield

Theoretical yield - the amount of product that could form during a reaction; it is calculated from a balanced chemical equation and it represents the maximum amount of product that could be formed from a given amount of reactant. Actual yield - the amount of product that forms when a reaction is carried out in the laboratory. It is measured in the lab. Why is the actual yield different from the percent yield? Percent yield - the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percent; it is a measure of the efficiency of a reaction

Slide 90 / 109 Theoretical Yield and % Yield

Theoretical Yield: Maximum amount of product that could be made. Limited by the amount of the limiting reactant. % Yield: The ratio of actual amount produced in the laboratory to the theoretical amount that could have been produced. Expressed as: Actual Yield x 100 Theoretical Yield

Slide 91 / 109 Calculating the Theoretical Yield

Example: Find the theoretical yield (in g) of AlCl3, if 27g Al and 71g Cl2 react. 2Al(s) + 3Cl2(g) --> 2AlCl3(s) Step 1: Determine the limiting reactant

27 g Al x 1 mol = 1 mol Al 71 g Cl2 x 1 mol = 1 mol Cl2 27 g 71 g 1 mol Al/2 = 0.50 mol Al available 1 mol Cl2/3 = 0.33 mol Cl2 available Cl2 Limits Step 2: Use INITIAL amount of Cl2 and stoichiometry to determine yield of desired product. 1 mol Cl2 x 2 mol AlCl3 x 133 g AlCl3 = 87.8 g 3 mol Cl2 1 mol AlCl3

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The efficiency of a reaction can be expressed as a ratio of the actual yield to the theoretical yield. For example, a percent yield of 85% shows that the reaction conditions are more favorable than with a percent yield of only 55%. Percent yield is the ratio comparing the amount actually obtained (actual yield) to the maximum amount that was possible (theoretical yield). Percent Yield = x 100

Percent Yield

Actual Yield Theoretical Yield

Slide 93 / 109 Calculating Theoretical Yield and % Yield

Example: A student burns 24 grams of methane with 30 L of

• xygen gas in the laboratory and produces 12.1 L of carbon

dioxide gas at STP. What is the % yield? CH4(g) + 2O2(g) --> CO2(g) + 2H2O(g)

Slide 94 / 109 Calculating Theoretical Yield and % Yield

Step 1: Find the LR 24 g CH4 x 1 mol = 1.5 mol CH4 30 L O2 x 1 mol = 1.33 mol O2 16 g 22.4 L 1.5 mol CH4/1 = 1.5 mol CH4 av. 1.33 mol O2/2 = 0.67 mol O2 av. O2 is LR Step 2: Find the theoretical yield (in L) using INITIAL amount of oxygen gas 1.33 mol O2 x 1 mol CO2 x 22.4 L = 19.8 L CO2 2 mol O2 1 mol Step 3: Calculate the % Yield 12.1 L CO2 Actual yield x 100 = 61.1% yield 19.8 L CO2 Theoretical yield

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45 What is the theoretical yield of phosphorus pentachloride if 2 grams of phosphorus trichloride react with 1.5 moles of chlorine gas @STP?

PCl3(g) + Cl2(g) --> PCl5(g) Slide 95 (Answer) / 109

45 What is the theoretical yield of phosphorus pentachloride if 2 grams of phosphorus trichloride react with 1.5 moles of chlorine gas @STP?

PCl3(g) + Cl2(g) --> PCl5(g)

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Answer 3.04

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46 At STP, what volume of laughing gas (dinitrogen monoxide) will be produced from 50 grams of nitrogen gas and 75 grams of oxygen gas? Remember to first write a balanced equation.

Slide 96 (Answer) / 109

46 At STP, what volume of laughing gas (dinitrogen monoxide) will be produced from 50 grams of nitrogen gas and 75 grams of oxygen gas? Remember to first write a balanced equation.

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Answer 40.1

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47 How many atoms of silver will be produced when 100 grams of copper react with 200 grams of silver nitrate?

Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)

Slide 97 (Answer) / 109

47 How many atoms of silver will be produced when 100 grams of copper react with 200 grams of silver nitrate?

Cu(s) + 2 AgNO3(aq) --> Cu(NO3)2(aq) + 2 Ag(s)

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Answer 7.08 x 10

23

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48 In the thermite reaction, aluminum reacts with iron (III)oxide to produce aluminum oxide and solid iron. If, when 258 grams of Al react with excess rust to produce 464 grams of pure iron, what is the % yield? (Remember, first write a balanced equation)

Slide 98 (Answer) / 109

48 In the thermite reaction, aluminum reacts with iron (III)oxide to produce aluminum oxide and solid iron. If, when 258 grams of Al react with excess rust to produce 464 grams of pure iron, what is the % yield? (Remember, first write a balanced equation)

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Answer 87%

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49 If 34 grams of ethane react with 84 L of oxygen gas to produce an actual yield in the laboratory of 2.9 moles of water vapor, what is the % yield? 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)

Slide 99 (Answer) / 109

49 If 34 grams of ethane react with 84 L of oxygen gas to produce an actual yield in the laboratory of 2.9 moles of water vapor, what is the % yield? 2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)

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Answer 91%

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50 Given the equation below, how many liters of sulfur dioxide would be actually produced if 55 grams of zinc sulfide were reacted with excess oxygen @STP and produced a 75% yield? 2ZnS(s) + 3O2(g) --> 2 SO2(g) + 2ZnO(s)

Slide 100 (Answer) / 109

50 Given the equation below, how many liters of sulfur dioxide would be actually produced if 55 grams of zinc sulfide were reacted with excess oxygen @STP and produced a 75% yield? 2ZnS(s) + 3O2(g) --> 2 SO2(g) + 2ZnO(s)

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Answer 9.5

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Calculating Excess Reactants

Slide 102 / 109 Calculating Excess Reactant

There will always be a certain amount of excess reactant

• remaining. The following steps are useful in determining how much
• f the excess reactant is left over.

Step 1: Use the limiting reactant to determine how much of the excess reactant was required to react. Step 2: Subtract the amount of excess reactant used from the initial amount.

Slide 103 / 109 Calculating Excess Reactant

Example: If 6 grams of hydrogen gas react with 160 grams of

• xygen gas, how much of the excess reactant remains?

2H2(g) + O2(g) --> 2H2O(g)

Slide 104 / 109

Step 1: Find the LR 6 g H2 x 1 mol H2 = 3 mol H2 160 g O2 x 1 mol = 5 mol O2 2 g H2 32 g O2 2 mol H2/2 = 1 mol H2 av. 5 mol/1 = 5 mol O2 av. H2 Limits Step 2: Use LR to find how much oxygen will be required. 3 mol H2 x 1 mol O2 = 1.5 mol O2 required 2 mol H2 Step 3: Subtract required amount from initial amount. 5 mol O2 initial - 1.5 mol O2 required = 3.5 mol Excess

2H2(g) + O2(g) --> 2H2O(g)

Calculating Excess Reactant Slide 105 / 109

51 How many grams of the excess reactant remain if 400 grams of nitrogen gas are reacted with 800 grams of oxygen gas according to the reaction below? (Don't forget to balance first!) ___N2(g) + ___O2(g) --> ___N2O5(g)

Slide 105 (Answer) / 109

51 How many grams of the excess reactant remain if 400 grams of nitrogen gas are reacted with 800 grams of oxygen gas according to the reaction below? (Don't forget to balance first!) ___N2(g) + ___O2(g) --> ___N2O5(g)

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Answer 120.4

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52 Methanol (CH3OH) can be synthesized from carbon monoxide and hydrogen gas. If 152 kg of carbon monoxide gas is reacted with 1500 L of H2 gas @STP, how many liters of the excess reactant remain? (Remember to first write a balanced reaction!)

Slide 106 (Answer) / 109

52 Methanol (CH3OH) can be synthesized from carbon monoxide and hydrogen gas. If 152 kg of carbon monoxide gas is reacted with 1500 L of H2 gas @STP, how many liters of the excess reactant remain? (Remember to first write a balanced reaction!)

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Answer 120,850

Slide 107 / 109

Credit to Tom Greenbowe Chemical Education Group at Iowa State University

Slide 108 / 109

Calcium hydroxide, Ca(OH)2, is also known as “slaked lime” and it is produced when water reacts with “quick lime,” CaO. If you start with 2.4 kg of quick lime, add excess water, and produce 2.06 kg of slaked lime, what is the percent yield of the reaction?

Stoichiometry Practice Problem

Is this a limiting reagent problem? Is the 2.06 kg a theoretical yield or actual yield? What quantity must you solve for? Did you write a balanced equation?

Slide 109 / 109