Section 12.2: Using Moles

• Used to relate moles of one substance to moles of another substance
• Use coefficients in the equation to convert to moles of other reactants or products

The following recipe serves 4:
4 potatoes (.3 lb each)
2 onions (.2 lb each)
8 carrots (.1 lb each)
4 stalks of celery (.05 lb each)
1.4 lbs of water
What is the total mass of the soup?
How would you make enough for 8?
How about 1240?
**With a balanced chemical equation and number of moles, we can predict the exact amount of reactant and product in a reaction**

There are 4 steps to follow:
1) Write the balanced chemical equation
2) Convert the given mass or volume to moles
3) Use the coefficients in the chemical equation to set up a mole ratio (the coefficients are the # of moles!)
HINT: The substance you are solving for goes on TOP
4) Convert these moles back to mass or volume as required

Use:
Grams A ↔ Moles A ↔ Moles B ↔ Grams B molar mass coefficients molar mass

N
2
+ 3H
2
 2NH
3
• This reaction can be stated as: 1 mole of nitrogen will react with 3 moles of hydrogen to produce 2 moles of ammonia.

#10) What mass of CO
2 forms when 95.6 g of C
3
H
8
C
3
H
8
(g) + 5O
2
(g) → 3CO
2
(g) + 4H
2
O(g) burns?
3 C = 36.033 g C= 12.011g
8 H= 8.064 g
C
3
H
8
2 O = 32 g
= 44.097 g/mol CO
2
= 44.011 g/mol
95.6 g C
3
H
8 x 1 mol C
3
H
8 x 3 mol CO
44.097g C
3
H
8
1 mol C
3
H
2
8 x 44.011 g CO
2
=
1 mol CO
2
= 286.2 g CO
2

#11) How many grams of fluorine are required to produce 10 g
XeF
6
?
Xe (g) + 3F
2
(g) → Xe F
6
(s)
F
2
: 2 x 18.998 g = 37.996 g
Xe F
6
: 131.29 + (6 x 18.998) = 245.278 g
10 g XeF
6 x 1 mol XeF
6
245.278 g XeF
6 x 3 mol F
2 x 37.996 g F
2
=
1 mol XeF
6
1 mol F
2
= 4.65 g F
2

The reactant that runs out first in a reaction/ stops the reaction

#1) What is the limiting reactant in producing water (H
H
2 or 5 g of O
2
2
? (convert g reactant → mol of product)
O), 5 g
2H
2
+ O
2
→ 2H
2
O
5 g H
2 x 1 mol H
2
2.0158 g H
2 x 2 mol H
2
O = 2.48 mol H
2
O
2 mol H
2
5 g O
2 x 1 mol O2 x 2 mol H2O = 0.31 mol H
2
O
31.998 g O2 1mol O
2
Limiting reactant is O
2
(runs out first)

2) Which is the limiting reactant in producing NH
3 g N
2 or 3.75 g H
2
?
, 3.75
N
2
+ 3H
2
→ 2 NH
3
3.75g H
2 x 1 mol H
2
2.0158 g H
2 x 2 mol NH
3
3 mol H
2
= 1.24 mol NH
3
3.75g N
2 x 1 mol N
2
28 g N
2 x 2 mol NH
3
1 mol N
2
= 0.27 mol NH
3
Limiting reactant is N
2