3.3.6 A Problem Using the Combined Concepts of Stoichiometry

Question 1

A Problem Using the Combined Concepts of Stoichiometry

Answer 1

• Several tools can be brought together to solve a stoichiometry problem

• Problem: 7.75 g of aluminum, 8.00 g of sodium hydroxide, and 100 g of water were reacted. 0.541 g of hydrogen gas was collected. What was the percent yield?
• First, the chemical equationmust be balanced. In the balanced chemical equation, 2 mol Al react with 2 mol NaOH and 6 mol H2O to produce 2 mol NaAl(OH)4 and 3 mol H2.
• Second, the massesof reagentsgiven must be converted to moles. The mass of each reactant is divided by its molar mass to give moles.
• For example, 7.75 g Al is equivalent to 0.287 mol Al.
• Next, the limiting reagent must be determined.
• To determine the limiting reagent, calculate how much hydrogen gas could be produced from each reactant. The reactant that results in the lowest number of moles of hydrogen gas is the limiting reagent. In this problem, NaOH is the limiting reagent.
• Next, the theoretical yieldmust be determined. Using the reactants given, 0.300 mol H2 can be produced. This is the theoretical yield of hydrogen gas.
• Finally, the percent yield must be calculated. The percent yield is the actual yielddivided by the theoretical yield and multiplied by 100%. In this problem, the actual yield is 0.541 g H2 ·1 mol H2 / 2.0158 g H2 = 0.268 mol H2. The percent yield is 89.3%. This percent yield means that 89.3% of the hydrogen gas that could have been produced was collected.

Question 2

If you have 5.0 mol Mg3N2, and 6.0 mol H2O, how many moles of Mg(OH)2 would be produced?

Mg3N2 + H2O -> Mg(OH)2 + NH3

This equation might need balancing.

Answer 2

3.0 mol Mg(OH)2

Question 3

What mass of AlCl3 can be formed from 5.000 mol NiCl2 and 2.000 mol Al?

Al(s) + NiCl2(aq) → AlCl3(aq) + Ni(s)

(Is the equation balanced?)

Answer 3

266.8 g AlCl3

Question 4

When aluminum foil is added to a solution of CuCl2, red copper metal starts to form on the foil. How many moles of Cu can be formed from 5 mol CuCl2 and 2 mol Al?

Al(s) + CuCl2(aq) → AlCl3(aq) + Cu(s)
(note that the reaction is not balanced!)

Answer 4

3 mol Cu

Question 5

If you start with 3.0 g of Cs and excess Cl2,
and obtain 3.0 g of CsCl,
what is the percent yield?

2Cs + Cl2 → 2CsCl

Answer 5

79%

Question 6

If you mix 12 moles of H2 and 7 moles of O2 to form H2O, which is the limiting reagent and why?

Answer 6

H2, because all 12 moles are used up.

Question 7

If the percent yield is 83%, what is the amount of MnBr2 that you would get if you add 3.00 g Mn to an excess of HBr?

Mn(s) + HBr(aq) → MnBr2(aq) + H2(g)

Answer 7

9.73 g MnBr2

Question 8

What mass of Ag2CO3 could you produce from 12.7 g AgNO3 assuming that it is the limiting reagent?

AgNO3(aq) + Na2CO3(aq) → Ag2CO3(aq) + NaNO3(aq)

This equation might need balancing.

Answer 8

10.3 g Ag2CO3

Question 9

How much CuSO4 is needed to react with an excess of aluminum to provide 14.0 g of Cu and any amount of aluminum sulfate?

CuSO4(s) + Al(s) → Cu(s) + Al2(SO4 )3(s)

Answer 9

35.2 g CuSO4

Question 10

How many moles of Li3N can be formed from 4.0 mol Li and 1.5 mol N2? (Don’t forget to balance the equation.)

Li + N2 -> Li3N

Answer 10

1.3 mol Li3N

Question 11

What is the maximum amount of K2SO4 that can be produced from 4.1 g of KOH according to the reaction below (which needs balancing)?

SO3 + KOH -> K2SO4 + H2O

Answer 11

6.4 g K2SO4