## Multiple Choice and Short Answer

Problem One
In the oxidation of ethane:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
how many moles of O2 are required to react with 1 mole of ethane?

Incorrect
This is the coefficient for O2, but the mole ratio of ethane to O2 is 7/2.

Incorrect
This is the coefficient for ethane, but it is the mole ratio of O2 to ethane that is important.

Correct
1 mole C2H6 x (7 mole O2/2 mole C2H6) = 7/2 mole O2

Problem Two
In the reaction:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
how many moles of CO2 are formed when 1 mole of O2 is consumed?

Incorrect
This is the coefficient of O2, but we need the relationship between the coefficients of CO2 and O2.

Incorrect
7 moles of O2 form 4 moles of CO2. Thus, the number of moles of CO2 must be smaller.

Correct
1 mole O2 x (4 mole CO2 / 7 mole O2 ) = 4/7 mole CO2

Problem Three
In the reaction:
2 C2H6 + 7 O2 → 4 CO2 + 6 H2O
how many moles of CO2 are formed when 5 moles of ethane are consumed?

Correct
5 mole C2H6 x (4 mole CO2 / 2 mole C2H6) = 10 mole CO2

Incorrect
Look at the ratio of coefficients of CO2 and ethane.

Incorrect
Look at the ratio of coefficients of CO2 and ethane.

Problem Four
How many mL of 0.1 M HCl are required to react with 0.01 mole of Na2CO3 (the products are NaCl and H2CO3)?

Incorrect
100 mL provides 0.01 mole, but the equation:
2 HCl + Na2CO3 → H2CO3 + 2 NaCl
indicates that 0.02 mole of HCl will be required.

Correct
200 mL provides 0.02 mole, required by the equation:
2 HCl + Na2CO3 → H2CO3 + 2 NaCl

Incorrect
100 mL provides 0.01 mole, but the equation:
2 HCl + Na2CO3 → H2CO3 + 2 NaCl
indicates that 0.02 mole of HCl will be required.

Problem Five
For the reaction:
2 Al + 6 HCl → 2 AlCl3 + 3 H2
how many liters of H2 at STP will be produced when 2.7 g Al and 100 mL of 1 M HCl are mixed (the reaction goes to completion)?

Incorrect
You probably used Al as the limiting reagent. Notice that three times more moles of HCl are required than Al.

Incorrect
You probably assumed a 1 to 1 mole ratio between reactants and products.

Correct
2.7 g Al is 0.10 mole, 100 mL of 1 M HCl is 0.10 mole. The equation requires 3 moles of HCl for every mole of Al. Thus, HCl is the limiting reagent.
0.10 mole HCl x (3 mole H2 / 6 mole HCl) = 0.05 mole H2
0.05 mole H2 x 22.4 L/mole at STP = 1.1 L

Problem Six
How many liters of H2 at STP are required to react with 2.3 g of Fe3O4 to give FeO and H2O?

Correct
The equation:
Fe3O4 + H2 → 3 FeO + H2O
shows a one to one mole ratio between Fe3O4 and H2. 2.3 g of Fe3O4 is 0.01 mole (molar mass = 231 g/mole) and therefore 0.01 mole of H2 is required. At STP 0.01 mole is 0.01 x 22.4 L/mole = 0.22 L.

Incorrect

Incorrect

Problem Seven
When 0.05 mole H2 is mixed with 0.05 mole CO, what is the maximum number of moles of methanol (CH3OH) that can be obtained (CH3OH is the only product)?

Incorrect

Incorrect

Correct
The equation:
2 H2 + CO → CH3OH
shows that 0.05 mole of CO would require 0.10 mole of H2. But only 0.05 mole H2 is present and produces only 0.025 mole CH3OH.

Problem Eight
When 10 g of copper and 10 g of iodine are mixed, what is the theoretical yield of CuI?
2 Cu + I2 → 2 CuI

Incorrect
You have probably used Cu as the limiting reagent, but it is in excess.

Correct
There is 0.039 mole of I2 and 0.16 mole Cu. Clearly, I2 is the limiting reagent.
0.039 mole I2 x (2 mole CuI / 1 mole I2 ) = 0.078 mole CuI, or 15 g

Incorrect
You have probably used a one to one mole ratio between I2 and CuI

Problem Nine
When 10 g of copper and 10 g of iodine are mixed, 5 g of CuI are obtained, what is the percent yield of CuI?
2 Cu + I2 → 2 CuI

Correct
As shown in the previous problem the theoretical yield is 15 g. If 5 g are actually obtain the percent yield is (5 g / 15 g) x 100 = 33%.

Incorrect
See the previous problem.

Incorrect
See the previous problem.

Problem Ten
A 0.4020 g sample of pure Na2C2O4 was titrated with 30.55 mL of a solution of NaMnO4, according to the equation:
2 NaMnO4 + 5 Na2C2O4 + 8 H2SO4 → 6 Na2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O
What was the molarity of the solution?

Incorrect
Check the coefficients in the equation.

Incorrect
Check the coefficients in the equation.

Correct
(0.4020 g/134.0 g/mole) = 3.000 x 10-3 moles Na2C2O4
3.000 x 10-3 moles x (2 NaMnO4 / 5 Na2C2O4) = 1.200 x 10-3 moles NaMnO4
(1.200 x 10-3 moles NaMnO4 / 0.03055 L) = 0.03928 M

Problem Eleven
A 5.104 g sample of impure Na2C2O4 was titrated with 30.55 mL of a 0.03928 M solution of NaMnO4, according to the equation:
2 NaMnO4 + 5 Na2C2O4 + 8 H2SO4 → 6 Na2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O
What is the percentage of Na2C2O4 in the sample?

Correct
A look at the previous question will show that there is a 5 to 2 mole ratio of Na2C2O4 to NaMnO4 . The 30.55 mL of 0.03928 M NaMnO4 contains:
0.03055 L x 0.03928 mole/L = 1.200 x 10-3 moles NaMnO4
which will react with:
1.200 x 10-3 moles x (5 mole Na2C2O4 / 2 mole NaMnO4) = 3.000 x 10-3 mole Na2C2O4
3.000 x 10-3 mole Na2C2O4 x 134.0 g/mole = 0.4020 g Na2C2O4
(0.4020 g Na2C2O4 / 5.104 g sample) x 100 = 7.876% Na2C2O4 per sample

Incorrect
See (a).

Incorrect
See (a).

Problem Twelve
The concentration of a NaMnO4 solution was determined by titrating 0.4020 g of pure Na2C2O4 with the solution. (This process is called standardizing the solution.) A 30.55 mL sample of the NaMnO4 was required to reach the endpoint. What is the concentration of the NaMnO4?

Incorrect
See problem 11.

Correct
This question is the reverse of the previous one. The sample of pure Na2C2O4 contains 0.4020 g/(134.0 g/mole) = 3.000 x 10-3 mole. According to the equataion:
2 NaMnO4 + 5 Na2C2O4 + 8 H2SO4 → 6 Na2SO4 + 2 MnSO4 + 10 CO2 + 8 H2O
3.000 x 10-3 mole Na2C2O4 x (2 NaMnO4 / 5 Na2C2O4) = 1.200 x 10-3 mole NaMnO4
(1.200 x 10-3 mole NaMnO4 / 0.03055 L) = 0.03928 M NaMnO4

Incorrect
See problem 11.

Problem Thirteen
What is the coefficient in front of H2 in the equation for the reaction of aluminum with HCl (the products are H2 and AlCl3)?

3

Problem Fourteen
What is the coefficient in front of benzene (C6H6) in the equation for the combustion of benzene (CO2 and water are the products)?

2

Problem Fifteen
What is the mole-mole ratio of tin to I2 in the preparation of SnI4 from tin and I2?

1 to 2

Problem Sixteen
What is the mole-mole ratio of acetylene to CO2 in the combustion of acetylene (C2H2)?

1 to 2

Problem Seventeen
How many moles of NO are formed from 5 moles of O2 in the oxidation of ammonia (NH3) to give NO and water?

4

Problem Eighteen
How many moles of NaOH are required to form 0.1 mole of Al(OH)3 according to the equation:
3 NaOH + AlCl3 → Al(OH)3 + 3 NaCl

0.3 moles

Problem Nineteen
One of the largest uses of sulfuric acid is in the production of phosphate fertilizers. When sulfuric acid (H2SO4) reacts with calcium phosphate (Ca3(PO4)2), calcium sulfate and calcium dihydrogenphosphate (Ca(H2PO4)2) are obtained. What mass of calcium sulfate is obtained from 1.0 x 105 moles of calcium phosphate?

2.7 x 107 g

Problem Twenty
If magnetite (an iron ore, Fe3O4) reacts completely with carbon to form CO2 and iron metal, how much magnetite must be used to obtain 1 pound of iron?

623 g

Problem Twenty One
If 1.0 kg of magnetite and 200 g carbon are mixed, how many moles of iron could be obtained in the reaction?

13 moles

Problem Twenty Two
If 10 g of magnesium and mixed with 2.0 g of Fe2O3, how many grams of Fe could be obtained in the reaction:
3 Mg + Fe2O3 → 2 Fe + 3 MgO

1.5 g

Problem Twenty Three
Spacecrafts and submarines use LiOH to capture exhaled CO2 in the reaction:
2 LiOH + CO2 → Li2CO3 + H2O
If one person exhales about 1000 g CO2 per day, how much LiOH is required to absorb that amount?

1090 g

Problem Twenty Four
If the reaction of copper metal with elemental iodine (I2) has a 60% yield, how much CuI, which is the only product, will be formed from a mixture of 2.0 g Cu and 2.0 g I2?

1.8 g

Problem Twenty Five
If the reaction of copper metal with elemental iodine (I2) has a 100% yield, how much I2 must be mixed with 2.0 g Cu in order to obtain 0.031 moles of CuI?

4.0

Problem Twenty Six
In the reaction of copper metal with I2 which produces only CuI, how much copper must be mixed with 127 g I2 in order to obtain 0.031 moles of CuI if the reaction has a 32% yield?

6.2 g

Problem Twenty Seven
One of the precursors to silicone polymers is prepared by heating methyl chloride with elemental silicon. If the reaction proceeds as follows:
2 CH3Cl + Si → (CH3)2SiCl2
and the yield is 75%, how much methyl chloride must be mixed with 100 g Si in order to obtain 0.10 mole of product?

13.5 g

Problem Twenty Eight
A 1.032 g sample containing sulfate ion is treated with excess BaCl2 solution. After "digestion", which allows the particles of BaSO4 to grow larger, the precipitate of BaSO4 is filtered, dried and weighed. From the weight of the precipitate, 0.4508 g, calculate the percent sulfate in the sample.

17.9%

Problem Twenty Nine
A 0.3456 g sample containing copper is dissolved in nitric acid, made basic with NaOH, and the resulting precipitate is filtered. The precipitate is then heated strongly to produce CuO, which weighs 0.2876 g. Calculate the percent copper in the sample.

66.48%

Problem Thirty
A stock solution of HCl is standardized by titrating a pure 0.3056 g sample of dry sodium carbonate. The titration required 30.20 mL of the HCl solution. Calculate the concentration of the solution.

0.1909 M

Problem Thirty One
A 1.5320 g sample of an impure carbonate is titrated with 0.09870 M HCl. 40.21 mL of the titrant are required to reach the endpoint. Calculate the percent carbonate in the sample.