Stoichiometry - AP level
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Example #1: A mixture of Mg and Zn with a combined mass of 1.0875 g was burned in oxygen producing MgO and ZnO with a combined mass of 1.4090 g. How many grams of zinc was in original mixture?

Solution #1:

1) Write the chemical equations:

2Zn + O2 ---> 2ZnO
2Mg + O2 ---> 2MgO

2) Assign a variable:

Let z be the mass of zinc to be found.
Then 1.0875 − z is the mass of Mg in the original mixture.

3) Zinc:

z   2 mol   81.3794 g  
––––––––––––  x  –––––  x  ––––––––  = 1.244676z
65.3820 g/mol   2 mol   1 mol  

In other words, the amount of zinc oxide produced is 1.244676 times the mass of zinc in the original mixture.

4) Magnesium:

1.0875 − z   2 mol   40.3044 g  
–––––––––––––  x  –––––  x  ––––––––  = 1.80337 − 1.65827z
24.30506 g/mol   2 mol   1 mol  

In other words, the amount of magnesium oxide produced is 1.80337 minus 1.65827 times the mass of magnesium in the original mixture.

5) Add the two masses of oxides and set the sum equal to the given total mass:

(1.244676z) + (1.80337 − 1.65827z) = 1.4090 g

z = 0.9535 g (of zinc)

Solution #2:

1) Let us assign the mass of Mg as x and the mass of Zn as y. We can write two simultaneous equations in two unknowns. The first is this:

x + y = 1.0875

2) The second is a bit more complex. Here it is:

(40.3044 / 24.30506) (x) + (81.3794 / 65.3820) (y) = 1.4090

3) Some explanation:

Write the equation for Mg becoming MgO:

Mg + 1⁄2O2 ---> MgO

We see that 1 mole of Mg will produce 1 mole of MgO. Converting to grams, we see that 24.30506 g of Mg will produce 40.3044 g of MgO. The producing of MgO from Mg will always be like this:

For any mass of Mg present, the amount of MgO produced will always be 40.304406/24.305 the amount of Mg originally present.

4) Solve:

1.658272x + 1.244676y = 1.4090

Eliminate the y:

1.658272x + (1.244676) (1.0875 − x) = 1.4090

1.658272x + 1.353585 − 1.244676x = 1.4090

0.413596x = 0.055415

x = 0.134 g

Example #2: A metal container of a fixed volume contains water vapor at 200. °C at a pressure of 4.50 atm. The water is then electrolytically split to generate hydrogen (H2) and oxygen (O2) gases. What will be the pressure (in atm) of the container at 200. °C after the water has been completely split?

Solution:

1) The balanced chemical equation is this:

2H2O(g) ---> 2H2(g) + O2(g)

2) Gay-Lussac's Law:

since the volume and temperature are constant, the amount (moles) of the gas and the pressure are directly proportional.

3) Therefore:

2   4.50 atm
–––  =  –––––––
3   x

x = 6.75 dm3

The three comes from the fact that every two volumes of water vapor produce three volumes (two of hydrogen gas and one of oxygen gas) of total gas.

Example #3: A metal container of fixed volume is filled with 2.50 atm of H2 gas and 2.50 atm of O2 at 200. °C. The gases then react to generate water vapor (H2O gas) through a combustion reaction. What is the total pressure (in atm) in the container at 200. °C when the reaction has progressed to the point where it has produced the maximum amount of H2O gas?

Solution:

1) Two preliminary points:

(a) this is a limiting reagent problem.
(b) because the temperature and volume are constant, the pressure and amount of gas (in moles) are in direct proportion.

2) The balanced chemical reaction:

2H2(g) + O2(g) ---> 2H2O(g)

3) Determine the limiting reagent:

H2 ---> 2.50 / 2 = 1.25
O2 ---> 2.50 / 1 = 2.50

Hydrogen gas is the limiting regent.

4) Determine the amount of water that is produced:

2   2.50 atm
–––  =  –––––––
2   x

x = 2.50 atm

5) Determine the amount of O2 that reacts:

2   2.50 atm
–––  =  –––––––
1   x

x = 1.25 atm

6) Determine the amount of O2 that is unreacted:

2.50 atm − 1.25 atm = 1.25 atm

7) Determine the total pressure in the container after reaction is complete:

2.50 atm + 1.25 atm = 3.75 atm

Example #4: 1 mole of hydrocarbon of formula CnH2n was burned completely in oxygen producing carbon dioxide and water vapour only. It required 192 grams of oxygen. Determine the formula of the hydrocarbon.

Solution:

1) Determine moles of O2 that combusted:

192 g / 32.0 g/mol = 6 mol
2) Balance the chemical equation of the reaction:
CnH2n + O2 ---> CO2 + H2O

CnH2n + O2 ---> nCO2 + nH2O

CnH2n + (3/2)nO2 ---> nCO2 + nH2O

2CnH2n + 3nO2 ---> 2nCO2 + 2nH2O

3) The mole ratio of CnH2n to O2 is 2 : 3n. Therefore:

2   1 <--- the one is from one mole of CnH2n given in the problem
–––  =  –––  
3n   6 <--- six moles of O2 required to combust one mole of CnH2n

3n = 12

n = 4

4) Formula of the hydrocarbon:

C4H8

Example #5: A 13.43 g mixture of CH4 and C2H6 is completely burned in excess oxygen. The mixture of CO2 and H2O that results weighs 64.84 g. Determine the fraction of CH4 in the original mixture.

Solution:

1) This problem requires two simultaneous equations in two unknowns. Here is the first equation:

x + y = 13.43

Where 'x' is the mass of CH4 in the mixture and 'y' is the mass of C2H6 in the mixture.

2) The second equation is this:

(80/16) (x) + (142/30) (y) = 64.84 g

The (80/16) and (142/30) will be explained!

3) The (80/16) factor:

Look at the balanced equation for the complete combustion of CH4:
CH4 + 2O2 ---> CO2 + 2H2O 
16 44 36 <--- that last is 2 moles of water

Look at the gram (not molar!) amounts of reactants and products:

16 grams of CH4 produce 44 g of CO2 plus 36 g of H2O

Therefore, 'x' grams of CH4 will produce (80/16) times 'x' grams of CO2 + H2O

4) The argument for the (142/30) factor is the same:

C2H6 + 3.5O2 ---> 2CO2 + 3H2O 
30 88 54 <--- 2 moles of carbon dioxide and 3 moles of water

'y' grams of C2H6 will produce (142/30) times 'y' grams of CO2 + H2O

5) Use y = 13.43 − x to eliminate y in the second equation:

(80/16) (x) + (142/30) (13.43 − x) = 64.84

Algebra!

5x + (4.733) (13.43 − x) = 64.84

5x + 63.57 − 4.733x = 64.84

0.267x = 1.27

x = 4.75655 g

6) The problem asked for the fraction of CH4. Let us assume that means mass percent:

(4.75655 g / 13.43 g) * 100 = 35.4%

Example #6: You are given 1.446 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl and O2:

2KClO3(s) ---> 2KCl(s) + 3O2(g)

and 355 mL of O2 is collected over water at 26.0 °C. The total pressure of the gases in the collection flask is 745.0 torr. What is the weight percentage of KClO3 in the sample?

Solution:

1) Use Dalton's Law to determine the pressure of the dry O2:

Ptotal = PO2 + PH2O

745.0 torr = PO2 + 25.2 torr

PO2 = 719.8 torr

2) Determine moles of O2 produced:

PV = nRT

(719.8 torr / 760.0 torr/atm) (0.355 L) = (n) (0.08206 L atm / mol K) (299 K)

n = 0.01370326 mol

3) Determine moles of KClO3 that decomposed. The KClO3 to O2 molar ratio is 2 to 3. Therefore:

2   x
–––  =  ––––––––––––––
3   0.01370326 mol

x = 0.0091355 mol

4) Determine mass of KClO3 and its percentage in the mixture:

(0.0091355 mol) (122.55 g/mol) = 1.120 g

(1.120 g / 1.446 g) (100) = 77.46%

Example #7: A mixture of potassium chloride and potassium bromide weighing 3.595 g is heated with chlorine gas, which converts the mixture completely to potassium chloride. The total mass of the KCl after the reaction is 3.129 g. What percent of the original mixture is KBr?

Solution:

1) We need two equations in two unknowns. The first is this:

x + y = 3.595 g

where 'x' is the mass of KCl and 'y' is the mass of KBr.

2) Some explanation:

(a) total moles present at end:
3.129 g / 74.55 g / mol = 0.04197 mol

(b) this is true:

Mole of KCl + Mole of KBr = 0.04197 mol

because the moles of KBr are replaced by moles of KCl in a 1:1 molar ratio.

3) Therefore, this is the second equation:

[x / 74.55 g/mol] + [y / 119.0 g/mol] = 0.04197 mol

4) Rearrange first equation, then substitute it into second equation (I dropped all units.):

y = 3.595 − x

[x / 74.55] − [(3.595 − x) / 119.0] = 0.04197

5) Solve:

[(x) (119.0)] + [(3.595 − x) (74.55)] = (0.04197) (119.0) (74.55)

119.0x − 74.55x + [(3.595) (74.55)] = 372.33

44.45x + 268.01 = 372.33

44.45x = 104.32

x = 2.347 g (of KCl)

y = 3.595 g − 2.347 g = 1.248 g (of KBr)

%KBr = (1.248 g / 3.595 g) (100) = 34.71%

Example #8: An alloy contain aluminum, copper, and zinc. You take a 12.657 g sample of the alloy and react it with an HCl solution. The reaction converts all of the aluminum and zinc in the alloy to aluminum chloride and zinc chloride in addition to producing hydrogen gas. The copper does not react with the HCl. Upon completion of the reaction, a total of 10.2 L of hydrogen gas was collected at a pressure of 726 torr and a temperature of 27.0 °C. Additionally, 2.642 g of unreacted copper is recovered. What is the percent composition of the alloy?

Solution:

1) Use PV = nRT:

(726 torr / 760 torr/atm) (10.2 L) = (n) (0.08206 liter-atm/mole K) (300.0 K)

n = 0.395795 mol

2) Determine combined grams of Al + Zn:

12.657 g minus 2.642 g = 10.015 g

3) The following balanced equations are required:

2Al + 6HCl ---> 2AlCl3 + 3H2

Zn + 2HCl ---> ZnCl2 + H2

4) Write a molar relationship between Al & Zn consumed and H2 produced:

the molar relationship between H2 produced and Al consumed is 3:2. That means this:
(3/2) (moles Al) = the moles of H2 produced by Al

In like manner:

(1/1) (moles Zn) = the moles of H2 produced by Zn

5) We know the total moles of H2 to be 0.395795 mol, which gives us this relationship:

[(3/2) (moles Al)] + [(1/1) (moles Zn)] = 0.395795 mol

6) Let y = grams Al. Then (10.015 − y) = grams Cu. Using the molar masses for Al and Zn gives us:

[(1.5) (y / 26.98)] + [(1/1) ((10.015 − y) / 65.409)] = 0.395795

0.055597y + 0.1531135 − 0.0152884y = 0.395795

0.0403086y = 0.2426815

y = 6.020 g <--- grams Al

grams Zn ---> 10.015 g − 6.020 g = 3.995 g

7) Let's make sure everything adds up:

2.642 g + 6.020 g + 3.995 g = 12.657 g

Yay!

Example #9: A sample of sodium bicarbonate (NaHCO3) is known to contain some impurities. It is found that Na makes up 18.00% of the entire mass of the sample. All of the Na comes from the NaHCO3. Find the mass percent of NaHCO3 in the sample.

Solution:

1) Let us determine the mass percent of Na in pure NaHCO3:

22.99 g / 84.0059 g = 0.27367

2) By how much of a factor would we have to increase the impurities to make the sodium percentage become 18%?

0.27367 / 0.18 = 1.52039

3) What is that increased amount in grams?

(84.0059 g) (1.52039) = 127.7 g

4) What is the mass percentage of the sodium bicarbonate in the 127.7 g sample?

84.0059 g / 127.7 g = 65.78%

Here's another way

Assume 84.0059 g of NaHCO3 is present in the sample.

This means 22.99 g of Na is present.

That 22.99 g represents 18% of the entire mixture.

What is the mass of the entire mixture?

22.99 is to 18% as x is to 100%

x = 127.7 g

The mass percent of NaHCO3 is:

84.0059 g / 127.7 g = 65.78%

Example #10: An ore of barium contains BaCO3. A 1.495 g sample of the ore was treated with HCl to dissolve the BaCO3. The resulting solution was filtered to remove insoluble material and then treated with H2SO4 to precipitate BaSO4. The precipitate was filtered, dried, and found to weigh 1.152 g. What is the percentage by mass of barium in the original sample?

Solution:

1) Molar mass of BaSO4:

233.391 g/mol

2) Moles of BaSO4 in the precipitate:

1.152 g / 233.391 g/mol = 0.0049359 moles

3) Moles of barium in the precipitate:

0.0049359 moles

By the 1:1 molar ratio between Ba and BaSO4

4) Mass of barium in the precipitate:

(0.0049359 mol) (137.327 g/mol) = 0.67783 g

5) Barium percent by mass in original sample:

(0.67783 g / 1.495 g) (100) = 45.34%

Problems #1-10

Problems #11-25

Problems #26-50

All Examples & Problems (no solutions)

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