Chapter 12: Chemical Quantities

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

Balanced chemical equations

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 equation2) Convert the given mass or volume to moles3) 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

Steps

N2 + 3H2 2NH3

•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 CO2 forms when 95.6 g of C3H8 burns?

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

3 C = 36.033 g C= 12.011g 8 H= 8.064 g 2 O = 32 gC3H8 = 44.097 g/mol CO2 = 44.011

g/mol

95.6 g C3H8 x 1 mol C3H8 x 3 mol CO2 x 44.011 g CO2 =

44.097g C3H8 1 mol C3H8 1 mol CO2

= 286.2 g CO2

Practice Problems (pg. 415)

#11) How many grams of fluorine are required to produce 10 g XeF6?

Xe (g) + 3F2 (g) → Xe F6 (s)

F2: 2 x 18.998 g = 37.996 g

Xe F6: 131.29 + (6 x 18.998) = 245.278 g

10 g XeF6 x 1 mol XeF6 x 3 mol F2 x 37.996 g F2 = 245.278 g XeF6 1 mol XeF6 1 mol F2

= 4.65 g F2

Practice (cont)

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

LIMITING REACTANT

#1) What is the limiting reactant in producing water (H2O), 5 g H2 or 5 g of O2? (convert g reactant → mol of product)

2H2 + O2 → 2H2O

5 g H2 x 1 mol H2 x 2 mol H2O = 2.48 mol H2O

2.0158 g H2 2 mol H2 5 g O2 x 1 mol O2 x 2 mol H2O = 0.31 mol H2O

31.998 g O2 1mol O2

Limiting reactant is O2 (runs out first)

Practice Problems

2) Which is the limiting reactant in producing NH3, 3.75 g N2 or 3.75 g H2?

N2 + 3H2 → 2 NH3

 

3.75g H2 x 1 mol H2 x 2 mol NH3 = 1.24 mol NH3

2.0158 g H2 3 mol H2  3.75g N2 x 1 mol N2 x 2 mol NH3 = 0.27 mol NH3

28 g N2 1 mol N2

Limiting reactant is N2

Practice (cont)