### 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 +3Cl_{2} → 2AlCl_{3}

a. 2 moles of Al will react with ______ mole(s) of Cl_{2} to produce ______ mole(s) of AlCl_{3}.

## Answer

2 moles of Al will react with 3 moles of Cl_{2} to produce 2 moles of AlCl_{3}

b. How many grams of AlCl_{3} will be produced if 2.50 moles of Al react?

## Answer

? g AlCl_{3} = 2.50 mol Al (2 mol AlCl_{3} / 2 mol Al) x (133.5 g AlCl_{3} / 1 mol AlCl_{3}) = 334 g AlCl_{3}

c. How many moles of Cl_{2} must react to produce 12.3 g of AlCl_{3}?

## Answer

? mol Cl_{2} = 12.3 g AlCl_{3} (1 mol AlCl_{3} / 133.5 g AlCl_{3}) x (3 mol Cl_{2} / 2 mol AlCl_{3}) = 0.138 mol Cl_{2}

d. How many grams of aluminum will react with 3.4 moles of chlorine?

## Answer

? g Al = 3.4 mol Cl_{2} (2 mol Al / 3 mol Cl_{2}) 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 AlCl_{3} = 17 g Al (1 mol Al / 27.0 g Al) x (2 mol AlCl_{3} / 2 mol Al) = 0.63 mol AlCl_{3}

2. The ammonia (NH_{3}) used to make fertilizers for lawns and gardens is made by reacting nitrogen and hydrogen according to the following reaction.

Balance the equation.

## Answer

N_{2} + 3H_{2} → 2NH_{3}

a. Determine the mass in grams of NH_{3} formed from 1.34 moles of nitrogen.

## Answer

? g NH_{3} = 1.34 mol N_{2} (2 mol NH_{3} / 1 mol N_{2}) x (17.0 g NH_{3} / 1 mol NH_{3}) = 45.6 g NH_{3}

b. What is the mass in grams of hydrogen required to react with 1.34 moles of nitrogen?

## Answer

? g H_{2} = 1.34 mol N_{2} (3 mol H_{2} / 1 mol N_{2}) x (2.0 g H_{2} / 1 mol H_{2}) = 8.06 g H_{2}

c. How many moles of nitrogen are required to produce 11.7 moles of NH_{3}?

## Answer

? mol N_{2} = 11.7 mol NH_{3} (1 mol N_{2} / 2 mol NH_{3}) = 5.85 mol N_{2}

d. How many moles of nitrogen are required to produce 11.7 grams of NH_{3}?

## Answer

? mol N_{2} = 11.7 g NH_{3} (1 mol NH_{3} / 17.0 g NH_{3}) x (1 mol N_{2} / 2 mol NH_{3}) = 0.344 mol N_{2}

e. How many grams of hydrogen are required to form 3.5 moles of NH_{3}?

## Answer

? g H_{2} = 3.5 mol NH_{3} (3 mol H_{2} / 2 mol NH_{3}) x (2.0 g H_{2} / 1 mol H_{2}) = 11 g H_{2}

3. The first step in the industrial manufacture of nitric acid involves the catalytic oxidation of ammonia according to the following BALANCED equation.

4NH_{3} (g) + 5O_{2} (g) 4NO (g) + 6H_{2}O (g)

a. How many moles of NO are formed if 824 g of NH_{3} react?

## Answer

? mol NO = 824 g NH_{3} (1 mol NH_{3} / 17.0 g NH_{3}) x (4 mol NO / 4 mol NH_{3}) = 48.5 mol NO

b. How many grams of water are formed if 2.5 moles of ammonia are oxidized?

## Answer

? g H_{2} = 2.5 mol NH_{3} (6 mol H_{2}O / 4 mol NH_{3}) x (18.0 g H_{2}O / 1 mol H_{2}O) = 68 g H_{2}O

c. How many moles of oxygen are needed to react with 4.6 moles of ammonia?

## Answer

? mol O_{2} = 4.6 mol NH_{3} (5 mol O_{2} / 4 mol NH_{3}) = 5.8 mol O_{2}

4. Mercury (II) oxide decomposes into mercury and oxygen gas according to the following UNBALANCED equation.

HgO Hg + O_{2}

a. How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?

## Answer

? mol HgO = 125 g O_{2} (1 mol O_{2} / 32.0 g O_{2}) x (2 mol HgO / 1 mol O_{2}) = 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 O_{2} = 2.3 mol Hg (1 mol O_{2} / 2 mol Hg) x (32.0 g O_{2} / 1 mol O_{2}) = 37 g O_{2}