1 CH 4 + H 2 O 3H 2 + CO If you have 32 g CH 4 and 32 g H 2 O, which - - PDF document

1

Unbalanced equation sample How many molecules of product can be formed with the sample given? Reactant A + Reactant B → Product C Balanced equation = A + B → C 2 sample Are any reactants leftover ? The reactant that determines the amount of product formed is called the limiting reactant Which reactant determined the amount of product formed?

Excess reactant

1 x B A 3 x C Reactant A + Reactant B → Products C + D + sample Balanced equation =A + B → C + D sample Are any reactants limiting ? Is anything leftover? + no Reactants A + B are in correct ratio as needed by balanced equation = the reactants are in a stoichiometric ratio 4 x C, 4 x D

CH4 + H2O 3H2 + CO

If you have 1 mole CH4 and 2 moles H2O, which is limiting reactant? If you have 1 mole CH4 and 2 moles H2O how many moles of CO can be produced? CH4 1 mole 2 mole 3 mole CH4 and 2 moles H2O how many moles of H2 can be produced? 6 mole

CH4 + H2O 3H2 + CO

3 mole CH4 and 2 moles H2O how many moles of CO can be produced?

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If you have 32 g CH4 and 32 g H2O, which is limiting? Convert to moles to ID LR: 32 g CH4 32 g H2O

• Will run out of H2O before CH4

Balanced eq. requires ratio

1 mole CH4 1 mole H2O

CH4 + H2O 3H

2 + CO

H2O = LR

4

1 mole CH4 16 g CH4

• = 2 mole CH4

O 1 mole H2O 18 g H2O

• = 1.77 mole H2O

Two types of LR problems:

• 1. Solving for amount of product formed

N2 + 3H

Step 1: Balance equation Step 2: Convert g to moles

32 g N2

• 5.92 g H2
• 32.0 g of nitrogen gas are reacted with 5.92 g

hydrogen gas to form ammonia. How much ammonia forms?

3H2 2NH 2NH3

2

1 mole N2 28 g N2

• = 1.14 mole N2

2

1 mole H2 2 g H2

• = 2.96 mole H2

2NH 3H

2.96 mole H2 1.14 mole N2

you have: You need:

H2 = LR

Step 3: Compare actual ratio (what you have) to what equation requires (what you need) to determine LR:

N2 + 3H2 2NH3

3 mole H2 1 mole N2 = 2.6 mole H2 1 mole N2

larger smaller simplify

Compare ratios Step 4: Use # moles of LR to perform stoichiometry (LR determines amount of product formed)

H2 = LR

N2 + 3H2 2NH3

2.96 mole H2

2

2 mole NH3 3 mole H2

• 17 g NH3

1 mole NH3

• = 33.5 g NH3
• 2. Solving for amount of excess

reactant remaining

3

26 g of oxygen are reacted with 3 g hydrogen to form water. How much of the excess reactant remains after the reaction is complete? Step 1: Balance equation Step 2: Convert g to moles

2H2 + O2 2H 2H2O

26 g O2

3 g H2

• 2

1 mole O2 32 g O2

• = 0.813 mole O2

2

1 mole H2 2 g H2

• = 1.5 mole H2

2H 2H

1.5 mole H2 0.813 mole O2

You have: You need:

2 mole H2 1 mole O2

H2 = LR

Step 3: Compare actual ratio you have to what equation requires to determine LR:

= 1.85 mole H2 1 mole O2

2H2 + O2 2H2O

Step 4: Use # moles of LR to perform stoichiometry to determine how much of

• ther reactant used in the reaction

2H2 + O2 2H2O

LR

• Calc. how much O2 consumed:

1.5 mole H2

2

1 mole O2 2 mole H2

• 32 g O2

1 mole O2

• = 24 g O2

Used in rxn

excess

LR moles

26 g

Starting mass O2 how much O2 used

• 24 g =

mass O2 remaining

2 g