Problems
It is very common for students to look at stoichiometry problems with the solution and think to themselves, “Yeah, I understand that. You set it up so that the units cancel. It makes perfect sense.” At this point, the student moves onto the next problem. When this student sees the same type of problem on a quiz or test, they are lost. Sure, it made perfect sense when they looked at the solution, but without actually trying to solve the problem themselves, nothing was learned. Do NOT look at the solution to these problems immediately. Actually try with a pencil, paper, calculator, and periodic table to solve these problems yourself. After you have tried, then look at the solution. Find your error(s). Learn from what you did wrong.
Check your solutions by clicking on the word Answer.
1. In a reaction between the elements aluminum and chlorine, aluminum chloride is produced.
Balance the equation.
Answer
2Al +3Cl2 → 2AlCl3
a. 2 moles of Al will react with ______ mole(s) of Cl2 to produce ______ mole(s) of AlCl3.
Answer
2 moles of Al will react with 3 moles of Cl2 to produce 2 moles of AlCl3
b. How many grams of AlCl3 will be produced if 2.50 moles of Al react?
Answer
? g AlCl3 = 2.50 mol Al (2 mol AlCl3 / 2 mol Al) x (133.5 g AlCl3 / 1 mol AlCl3) = 334 g AlCl3
c. How many moles of Cl2 must react to produce 12.3 g of AlCl3?
Answer
? mol Cl2 = 12.3 g AlCl3 (1 mol AlCl3 / 133.5 g AlCl3) x (3 mol Cl2 / 2 mol AlCl3) = 0.138 mol Cl2
d. How many grams of aluminum will react with 3.4 moles of chlorine?
Answer
? g Al = 3.4 mol Cl2 (2 mol Al / 3 mol Cl2) x (27.0 g Al / 1 mol Al) = 61.2 g Al
e. If 17 grams of aluminum react, how many moles of aluminum chloride will be produced?
Answer
? mol AlCl3 = 17 g Al (1 mol Al / 27.0 g Al) x (2 mol AlCl3 / 2 mol Al) = 0.63 mol AlCl3
2. The ammonia (NH3) used to make fertilizers for lawns and gardens is made by reacting nitrogen and hydrogen according to the following reaction.
Balance the equation.
Answer
N2 + 3H2 → 2NH3
a. Determine the mass in grams of NH3 formed from 1.34 moles of nitrogen.
Answer
? g NH3 = 1.34 mol N2 (2 mol NH3 / 1 mol N2) x (17.0 g NH3 / 1 mol NH3) = 45.6 g NH3
b. What is the mass in grams of hydrogen required to react with 1.34 moles of nitrogen?
Answer
? g H2 = 1.34 mol N2 (3 mol H2 / 1 mol N2) x (2.0 g H2 / 1 mol H2) = 8.06 g H2
c. How many moles of nitrogen are required to produce 11.7 moles of NH3?
Answer
? mol N2 = 11.7 mol NH3 (1 mol N2 / 2 mol NH3) = 5.85 mol N2
d. How many moles of nitrogen are required to produce 11.7 grams of NH3?
Answer
? mol N2 = 11.7 g NH3 (1 mol NH3 / 17.0 g NH3) x (1 mol N2 / 2 mol NH3) = 0.344 mol N2
e. How many grams of hydrogen are required to form 3.5 moles of NH3?
Answer
? g H2 = 3.5 mol NH3 (3 mol H2 / 2 mol NH3) x (2.0 g H2 / 1 mol H2) = 11 g H2
3. The first step in the industrial manufacture of nitric acid involves the catalytic oxidation of ammonia according to the following BALANCED equation.
4NH3 (g) + 5O2 (g) 
4NO (g) + 6H2O (g)
a. How many moles of NO are formed if 824 g of NH3 react?
Answer
? mol NO = 824 g NH3 (1 mol NH3 / 17.0 g NH3) x (4 mol NO / 4 mol NH3) = 48.5 mol NO
b. How many grams of water are formed if 2.5 moles of ammonia are oxidized?
Answer
? g H2 = 2.5 mol NH3 (6 mol H2O / 4 mol NH3) x (18.0 g H2O / 1 mol H2O) = 68 g H2O
c. How many moles of oxygen are needed to react with 4.6 moles of ammonia?
Answer
? mol O2 = 4.6 mol NH3 (5 mol O2 / 4 mol NH3) = 5.8 mol O2
4. Mercury (II) oxide decomposes into mercury and oxygen gas according to the following UNBALANCED equation.
HgO 
Hg + O2
a. How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?
Answer
? mol HgO = 125 g O2 (1 mol O2 / 32.0 g O2) x (2 mol HgO / 1 mol O2) = 7.81 mol HgO
b. How many moles of mercury are produced if 24.5 moles of mercury (II) oxide decompose?
Answer
? mol Hg = 24.5 mol HgO (2 mol Hg / 2 mol HgO) = 24.5 mol Hg
c. How many grams of oxygen will be produced if 2.3 moles of mercury are produced?
Answer
? mol O2 = 2.3 mol Hg (1 mol O2 / 2 mol Hg) x (32.0 g O2 / 1 mol O2) = 37 g O2