1. Aim # 12: How do we solve mass-mass problems?
HW # 12
Study pp pp
Study class notes
Ans. ques. p. 282 # 11
p. 283 # 17,18,21,23,42

2. I Solving mass-mass problems
Problem: Based upon the equation
3H2(g) + N2(g) → 2NH3(g)
How many grams of ammonia are produced by the reaction of 56 g of nitrogen with excess hydrogen?
Solution: For the reaction A+B→C+D
mass x moles x moles x mass
A ÷ A ÷ C ÷ C

3. Step 1. Change the given mass to moles using an appropriate conversion factor e.g. 56g N2 x 1 mol N2 = 2.0 mol N g N2
Step 2. Find the number of moles of the substance sought using coefficients from the balanced equation.
e.g. 2.0 mol N2 x 2 mol NH3 = 4.0 mol NH3
1 mol N2
Step 3. Change the moles of the substance sought to mass using an appropriate conversion factor.
e.g. 4.0 mol NH3 x 17 g NH3 = 68 g NH3
1 mol NH3

4. II Problem: How many grams of hydrogen gas are produced when enough nitric acid is added to g of magnesium to react it completely? Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)
Ans g Mg x 1 mol Mg = 2.00 mol Mg
24.3 g Mg
2.00 mol Mg x 1 mol H2 = 2.00 mol H2
1 mol Mg
2.00 mol H2 x 2.00 g H2 = 4.00 g H2
1 mol H2

5. III Percent Yield
Yield is the mass of product obtained.
% yield = actual yield x 100
theoretical yield
In the previous problem, if we obtained
3.70 g H2, what would our % yield be?
% yield = 3.70 g x 100 = 92.5% g

6. Problem: For the reaction C2H6 + 7O2 → 4CO2 + 6H2O How many grams of ethane (C2H6) need to be burned in order to produce 162 grams of water?
Ans g H2O x 1 mol H2O = 9.00 mol H2O
18.0 g H2O
9.00 mol H2O x 2 mol C2H6 = 3.00 mol C2H6
6 mol H2O
3.00 mol C2H6 x 30.0 g C2H6 = g C2H6
1 mol C2H6