LIMITING REAGENT Practice Problems



Chemistry I-H

Stoichiometry / Limiting Reagent Problem Set

Answer Key

1. At high temperatures, sulfur combines with iron to form the brown-black iron (II) sulfide:

Fe (s) + S (l) ( FeS (s)

In one experiment, 7.62 g of Fe are allowed to react with 8.67 g of S.

a. What is the limiting reagent, and what is the reactant in excess?

b. Calculate the mass of FeS formed.

SEE YOUR NOTES FOR THE ANSWERS

2. Arcylonitrile, C3H3N, is the starting material for the production of a kind of synthetic fiber acrylics) and can be made from propylene, C3H6, by reaction with nitric oxide, NO, as follows:

4 C3H6 (g) + 6 NO (g) → 4 C3H3N (s) + 6 H2O (l) + N2 (g)

What mass of C3H3N can be made when 21.6 g of C3H6 react with 21.6 g of nitric oxide?

THIS IS A LIMITING REACTANT PROBLEM BECAUSE WE HAVE AMOUNTS OF BOTH REACTANTS! SOLVE TWICE!

21.6 g C3H6 x (1 mole C3H6) x ( 4 mole C3H3N) x (53.07 g C3H3N) = 27.2 g C3H3N

(42.09 g C3H6) ( 4 mole C3H6) (1 mole C3H3N)

21.6 g NO x (1 mole NO) x ( 4 mole C3H3N) x (53.07 g C3H3N) = 25.5 g C3H3N

(30.01 g NO) (6 mole NO) (1 mole C3H3N)

THE ANSWER TO A LIMITING REACTANT STYLE QUESTION IS THE SMALLEST AMOUNT OF PRODUCT! = 25.5 g C3H3N

3. Calculate the percent yield for the reaction: P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) if 75.0 g of phosphorus reacts with excess chlorine gas to produce 111.0 g of phosphorus trichloride.

IF IT SAYS PERCENT YIELD, YOU KNOW THE AMOUNT OF 1 REACTANT AND THE ACTUAL YIELD OF 1 PRODUCT. START WITH REACTANT AND SOLVE FOR THEORETICAL YIELD OF PRODUCT, THEN SOLVE FOR PERCENT.

75.0 g P4 x (1 mole P4) x ( 4 mole PCl3) x (137.32 g PCl3) = 333 g PCl3 is the theoretical yield

(123.88 g P4) (1 mole P4) (1 mole PCl3)

Percent Yield = Actual Yield x 100 = 111.0 g PCl3 = 33.33% Yield

Theoretical Yield 333 g PCl3

4. Formic acid, HCHO2, burns in oxygen to form carbon dioxide and water as follows:

HCHO2 (aq) + O2 (g) → 2 CO2 (g) + 2 H2O (l).

If a 3.15-g sample of formic acid was burned in 2.0 L of oxygen, what volume of carbon dioxide would be produced? (Assume the reaction occurs at standard temperature and pressure, STP.)

THIS IS A LIMITING REACTANT PROBLEM BECAUSE WE HAVE AMOUNTS OF BOTH REACTANTS!

3.15 g HCHO2 x (1 mole HCHO2) x ( 2 mole CO2) x (22.4 L CO2) = 3.07 L CO2

(46.03 g HCHO2) (1 mole HCHO2) (1 mole CO2)

2.0L O2 x (1 mole O2) x ( 2 mole CO2) x (22.4 L CO2) = 4.0 L CO2

(22.4L O2) (1 mole O2) (1 mole CO2)

THE ANSWER TO A LIMITING REACTANT PROBLEM IS THE SMALLER VALUE = 3.07 L CO2

5. Zinc metal reacts with hydrochloric acid to produce zinc chloride and hydrogen gas.

a. Balance the following reaction: Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)

b. A 3.50-g sample of zinc metal is allowed to react with 2.50 g of hydrochloric acid.

Complete the following table:

|Reactants/products |Zn (grams) |HCl (grams) |ZnCl2 (grams) |H2 (L) |

|Before reaction |3.50 g |2.50 g |0g |0g |

|After reaction |1.26 g |0 g | | |

THE INFORMATION FILLED IN ABOVE CAN BE DETERMINED FROM READING THE QUESTION. READING THE TABLE CAREFULLY INDICATES THAT SINCE THERE IS LEFT OVER ZINC, HCl IS THE LIMITING REACTANT AND WE SHOULD CALCULATE OUR PRODUCTS FROM THAT NUMBER.

2.50 g HCl x (1 mole HCl) x ( 1 mole ZnCl2) x (136.28 g ZnCl2) = 4.67 g ZnCl2

(36.46 g HCl) (2 mole HCl) (1 mole ZnCl2)

2.50 g HCl x (1 mole HCl) x ( 1 mole H2) x (22.4L H2) = 0.768 L H2

(36.46 g HCl) (2 mole HCl) (1 mole H2)

6. Consider the reaction: MnO2 + 4 HCl → MnCl2 + Cl2 + 2 H2O

If 0.45 mols of MnO2 can react with 48.2 g of HCl, how many grams of Cl2 could be produced?

THIS IS A LIMITING REACTANT PROBLEM BECAUSE WE HAVE AMOUNTS OF BOTH REACTANTS!

0.45 moles MnO2 x ( 1 mole Cl2) x (70.90 g Cl2) =32 g Cl2

(1 mole MnO2) (1 mole Cl2)

48.2 g HCl x (1 mole HCl) x ( 1 mole Cl2) x (70.90 g Cl2) = 23.4 g Cl2

(36.46 g HCl) ( 4 mole HCl) (1 mole Cl2)

THE ANSWER TO A LIMITING REACTANT PROBLEM IS THE SMALLER OF THE TWO NUMBERS, ONCE THAT AMOUNT OF PRODUCT IS MADE, THE REACTION STOPS!

Answer = 23.4 grams made

7. One of the components of the fuel mixture on the Apollo lunar module involved a reaction with hydrazine, N2H4, and dinitrogen tetraoxide, N2O4. If the balanced equation for this reaction is

2 N2H4 (l) + N2O4 (g) → 3 N2 (g) + 4 H2O (g),

What volume of N2 gas (measured at STP) would result from the reaction of 1500 kg of hydrazine and 1000 kg of N2O4?

THIS IS A LIMITING REACTANT PROBLEM BECAUSE WE HAVE AMOUNTS OF BOTH REACTANTS!

We have to change kg to g first before we convert to moles.

1500 kg N2H4 x ( 103 g N2H4) x ( 1 mole N2H4) x( 3 mole N2) x (22.4 L N2) = 1.6x106 L N2

(1 kg N2H4) ( 32.06 g N2H4) (2 mole N2H4) (1 mole N2)

1000 g N2O4 x ( 103 g N2O4) x ( 1 mole N2O4) x( 3 mole N2) x (22.4 L N2) = 7.3 x 105 L N2

(1 kg N2O4) ( 92.02 g N2O4) (1 mole N2H4) (1 mole N2)

SMALLEST ANSWER = TRUE ANSWER = 7.3 x 105 L N2

8. Calculate the percent yield for an experiment in which 5.50 g of SOCl2 was obtained in a reaction of 5.80 g of SO2 with excess PCl5. Use the following equation:

SO2 (l) + PCl5 (l) → SOCl2 (l) + POCl3 (l).

IF IT SAYS PERCENT YIELD, YOU KNOW THE AMOUNT OF 1 REACTANT AND THE ACTUAL YIELD OF 1 PRODUCT. START WITH REACTANT AND SOLVE FOR THEORETICAL YIELD OF PRODUCT, THEN SOLVE FOR PERCENT.

5.80 g SO2 x (1 mole SO2) x ( 1 mole SOCl2) x (118.97 g SOCl2) = 10.8 g SOCl2 is theoretically produced

(64.07 g SO2) ( 1 mole SO2) (1 mole SOCl2)

Percent Yield = Actual Yield x 100 = 5.50 g SOCl2 x 100 = 50.9%

Theoretical Yield 10.8 g SOCl2

9. Chlorine gas reacts with silica, SiO2, and carbon to give silicon tetrachloride and carbon monoxide.

a. Balance the following equation: Cl2 (g) + SiO2 (s) + C (s) → SiCl4 (l) + CO (g)

b. How much CO gas can be produced from 15.0 g of silica?

WE ONLY KNOW ONE QUANTITY, SO START THERE AND SOLVE ONCE FOR WHAT EVER THE QUESTION ASKS FOR.

2 Cl2 (g) + SiO2 (s) + 2 C (s) → SiCl4 (l) + 2 CO (g)

15.0 g SiO2 x (1 mole SiO2) x ( 2 mole CO) x (28.01 g CO) = 14.0 g CO

(60.09 g SiO2) (1 mole SiO2) (1 mole CO)

10. When iron (II) hydroxide is mixed with phosphoric acid, iron (II) phosphate precipitate results.

a. Balance the following equation: 3Fe(OH)2 (aq) + 2H3PO4 (aq) → Fe3(PO4)2 (s) + 6H2O (l)

b. If 3.20 g of Fe(OH)2 is treated with 2.50 g of phosphoric acid, what is the limiting reagent and what is the reactant in excess?

Look at question C because you are asked for the mass of Fe3(PO4)2 there so solve for that in this question to save time.

3.20 g Fe(OH)2 x (1 mole Fe(OH)2) x ( 1 mole Fe3(PO4)2) x (357.49 g Fe3(PO4)2) = 4.24 g Fe3(PO4)2

(89.87 g Fe(OH)2) (3 mole Fe(OH)2) (1 mole Fe3(PO4)2)

2.50 g H3PO4 x (1 mole H3PO4) x ( 1 mole Fe3(PO4)2) x (357.49 g Fe3(PO4)2) = 4.56 g Fe3(PO4)2

(98.00 g H3PO4) (2 mole H3PO4) (1 mole Fe3(PO4)2)

Because Fe(OH)2 produces the least amount of product, it is the limiting reactant, and H3PO4 is the excess.

c. How many grams of Fe3(PO4)2 precipitate can be formed? The least amount of product is the answer(4.24 g Fe3(PO4)2

d. If 3.99 g of Fe3(PO4)2 is actually obtained, what is the percent yield?

Percent Yield = 3.99 g x 100 = 94.1 g Fe3(PO4)2

4.24 g

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