Differentiated+Stoichiometry+Questions

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 * TRUE/FALSE **

1. //Stoichiometry// is the study of the relationship between the amount of reactants used and the amount of products made in a chemical reaction.

ANS: T PTS: 1 **DIF: Bloom's Level 1|DOK 1**

2. In a balanced chemical equation, the total number of moles of all the reactants is equal to the total number of moles of all the products.

ANS: F PTS: 1 **DIF: Bloom's Level 2|DOK 1**

3. In a chemical reaction represented by the general equation, there are four distinct mole ratios that can be written.

ANS: F PTS: 1 **DIF: Bloom's Level 3|DOK 1**

4. The //mole ratio// is a comparison of how many moles of one substance are required to participate in a chemical reaction with another substance, based on the balanced chemical equation.

ANS: T PTS: 1 **DIF: Bloom's Level 2|DOK 1**

5. In a chemical reaction, the reactant with the largest molar mass is the limiting reagent.

ANS: F PTS: 1 **DIF: Bloom's Level 2|DOK 1**

6. The //excess reagent// refers to the additional amount of a reactant in a chemical reaction that must be added in order for the reaction to proceed.

ANS: F PTS: 1 **DIF: Bloom's Level 2|DOK 1**

7. The stoichiometric relationship between any two substances in a reaction depends on the mole ratio between those substances.

ANS: T PTS: 1 **DIF: Bloom's Level 2|DOK 1**

8. The //limiting reagent// limits the amount of product formed in a chemical reaction.

ANS: T PTS: 1 **DIF: Bloom's Level 1|DOK 1**

9. In a balanced chemical equation, the total number mass of the reactants is equal to the total mass of the products.

ANS: T PTS: 1 **DIF: Bloom's Level 2|DOK 1**

10. The percent yield is the maximum amount of product that can be produced from a given amount of reactant.

ANS: F The percent yield is the ratio of the actual yield to the theoretical yield, and is expressed as a percent.

PTS: 1 **DIF: Bloom's Level 1|DOK 1** TOP: Determine the percent yield for a chemical reaction. KEY: Theoretical yield

NOT: The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant.

11. If 80.0 g of sodium reacts with 72.0 g of iron (III) oxide, sodium acts as the limiting reactant.

ANS: F

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Identify the limiting reactant in a chemical equation. KEY: Limiting reactant

NOT: The actual ratio (0.129) is less than the required ratio (0.166). Thus, iron (III) oxide is the limiting reactant.


 * MULTIPLE CHOICE **

1. Based on the mole ratios of the substances in a chemical reaction shown, which is the correct equation for the chemical reaction?
 * Substances || Mole ratio ||
 * A:B || 3:2 ||
 * A:C || 3:1 ||
 * B:C || 2:1 ||


 * a. ||  || c. ||   ||
 * b. ||  || d. ||   ||

ANS: D PTS: 1 **DIF: Bloom's Level 2|DOK 2**

2. What is the mole ratio of all the substances represented in this figure?


 * a. || 4:6:10 || c. || 2:3:5 ||
 * b. || 2:3:2 || d. || 2:2:5 ||

ANS: B PTS: 1 **DIF: Bloom's Level 3|DOK 2**

3. Which is true of the reaction shown below?


 * a. || The mole ratio of this reaction is 6:5:6. ||
 * b. || Two molecules of Substance Y will be left over when this reaction goes to completion. ||
 * c. || Substance Y is the limiting reagent in this reaction. ||
 * d. || The addition of more molecules of Substance X will not affect the amount of Substance Z that can be made. ||

ANS: B PTS: 1 **DIF: Bloom's Level 4|DOK 2**

4. Which conversion factor will correctly complete this setup for finding the number of moles of O2 required to completely react with 75.0 grams of Sb? The balanced chemical equation for the reaction is shown:


 * a. || 1 mole/31.998 g O2 || c. || 3 moles O2/4 moles Sb ||
 * b. || 4 moles Sb/3 moles O2 || d. || 31.998g O2 /1 mole ||

ANS: C PTS: 1 **DIF: Bloom's Level 4|DOK 2**

5. The table shows the mole ratios of potassium and bromine combined to form potassium bromide according to the balanced reaction. In which trial(s) is the amount of bromine the limiting reagent?
 * Trial Number || Moles of K || Moles of Br ||
 * 1 || 10 || 3 ||
 * 2 || 3 || 2 ||
 * 3 || 14 || 8 ||


 * a. || Trials 1 and 3 || c. || Trials 2 and 3 ||
 * b. || Trial 2 only || d. || Trial 1 only ||

ANS: D PTS: 1 **DIF: Bloom's Level 4|DOK 2**

6. Gold is reacted with chlorine gas according to the reaction 2 Au + 3 Cl2 → 2 AuCl3. Use the data in the table to determine the percent yield of gold chloride.
 * Mass of Gold || Mass of Chlorine || Theoretical Yield || Actual Yield ||
 * 39.4 g || 21.3 g || ? || 35.2 g ||


 * a. || 60.7 % || c. || 58.1 % ||
 * b. || 54.1 % || d. || 51.4 % ||

ANS: C PTS: 1 **DIF: Bloom's Level 4|DOK 2**

7. According to this chemical reaction, which is the number of grams of Fe produced from 14 moles of H2? Fe3O4 (cr) + 4 H2 (g) → 3 Fe (cr) + 4 H2O (l)
 * a. || 587 g || c. || 4862 g ||
 * b. || 2431 g || d. || 22.1 g ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

8. Which is the correct mole ratio of K3PO4 to KNO3 in the chemical reaction Mg(NO3)2 + K3PO4 → Mg3(PO4)2 + KNO3?
 * a. || 1:1 || c. || 1:3 ||
 * b. || 2:3 || d. || 1:2 ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 1**

9. Which is the correct mole ratio for aluminum chloride to chlorine in the chemical reaction AlCl3 + Br2 → AlBr3 + Cl2?
 * a. || 1:1 || c. || 1:3 ||
 * b. || 2:3 || d. || 3:2 ||

ANS: B PTS: 1 **DIF: Bloom's Level 3|DOK 2**

10. How many moles of KBr will be produced from 7 moles of BaBr2? BaBr2 + K2SO4 → KBr + BaSO4
 * a. || 1 mole || c. || 14 moles ||
 * b. || 7 moles || d. || 3.5 moles ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2**

11. How many moles of Al would be produced from 20 moles of Al2O3? Al2O3 → Al + O2
 * a. || 10 moles || c. || 40 moles ||
 * b. || 20 moles || d. || 4 moles ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2**

12. How many moles of Cu are needed to react with 5.8 moles of AgNO3? Cu + 2 AgNO3 → Cu(NO3)2 + 2 Ag
 * a. || 2.9 moles || c. || 5.8 moles ||
 * b. || 3.8 moles || d. || 11.6 moles ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

13. Which is the number of moles of carbon dioxide produced from the complete combustion of 5.42 moles of ethanol? C2H6O + O2 → CO2 + H2O
 * a. || 5.42 moles || c. || 10.84 moles ||
 * b. || 2.71 moles || d. || 16.3 moles ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2**

14. Which is the correct number of moles of NO that is produced from 13.2 moles of oxygen gas in the presence of excess ammonia? 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
 * a. || 10.6 moles || c. || 1.52 moles ||
 * b. || 12.2 moles || d. || 16.5 moles ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

15. How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen? H2 + O2 → H2O
 * a. || 45.0 g || c. || 90.0 g ||
 * b. || 22.5 g || d. || 3.6 g ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2**

16. How many grams of O2 are required to produce 358.5 grams of ZnO? 2Zn + O2 → 2ZnO
 * a. || 1302 g || c. || 29.1 g ||
 * b. || 14.5 g || d. || 70.5 g ||

ANS: D PTS: 1 **DIF: Bloom's Level 3|DOK 2**

17. How many grams of bromine are required to react completely with 37.4 grams aluminum chloride? AlCl3 + Br2 → AlBr3 + Cl2
 * a. || 33.6 g || c. || 29.9 g ||
 * b. || 134.5 g || d. || 67.2 g ||

ANS: D PTS: 1 **DIF: Bloom's Level 3|DOK 2**

18. How many grams of chlorine gas can be produced from the decomposition of 73.4 g. of AuCl3 by this reaction: 2AuCl3 → 2 Au + 3 Cl2
 * a. || 25.7 g || c. || 11.4 g ||
 * b. || 33.6 g || d. || 195.3 g ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

19. A chemical reaction can theoretically produce 137.5 grams of product, but in actuality 112.9 grams are produced. Which is the percent yield for this reaction?
 * a. || 82.1% || c. || 24.6% ||
 * b. || 62.0 g || d. || 17.9% ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

20. How many moles of carbon dioxide are produced when 19.3 mol of propane gas is burned in excess oxygen?


 * a. || 6.43 mol || c. || 32.2 mol ||
 * b. || 57.9 mol || d. || 6.84 mol ||

ANS: B PTS: 1 **DIF: Bloom's Level 3|DOK 2**

21. What is the mass of potassium chloride when 6.75 g of potassium reacts with an excess of chlorine gas? The balanced chemical equation is: .
 * a. || 6.45 g || c. || 12.9 g ||
 * b. || 25.7 g || d. || 5.79 g ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2**

22. A certain reaction has a 73.6% yield. If 53.8 grams of the product were predicted by stoichiometry to be made, what would the actual yield be?
 * a. || 73.1 g || c. || 26.9 g ||
 * b. || 19.8 g || d. || 39.6 g ||

ANS: D PTS: 1 **DIF: Bloom's Level 3|DOK 2**

23. A reaction was predicted to produce 32.4 grams of a compound. When the product was measured, there were only 26.1 grams made. What is the percent yield of this reaction?
 * a. || 80.6% || c. || 24.1% ||
 * b. || 6.3% || d. || 58.5% ||

ANS: A PTS: 1 **DIF: Bloom's Level 3|DOK 2**

24. How many grams of H2O will be produced if 750 grams of Fe are produced?


 * a. || 17.9 g || c. || 563 g ||
 * b. || 322 g || d. || 241 g ||

ANS: B PTS: 1 **DIF: Bloom's Level 3|DOK 2**

25. How many grams of Fe3O4 are required to react completely with 300 grams of H2?


 * a. || 2096 g || c. || 8681 g ||
 * b. || 1.54 g || d. || 37.5 g ||

ANS: C PTS: 1 **DIF: Bloom's Level 3|DOK 2** NAT: UCP.3 | B.3

26. How many moles of carbon dioxide is produced when 10.4 mol of propane gas is burned in excess oxygen?
 * a. || 0.288 mol || c. || 31.2 mol ||
 * b. || 3.46 mol || d. || 52.0 mol ||

ANS: C The equation for the combustion of propane is.


 * || ** Feedback ** ||
 * ** A ** || Divide the unknown moles of carbon dioxide by the known moles of propane. ||
 * ** B ** || Multiply the known number of moles of propane by the mole ratio. ||
 * ** C ** || Correct! ||
 * ** D ** || Balance the equation correctly. ||

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Use the steps to solve stoichiometric problems. KEY: Stoichiometric mole-to-mole conversion

27. How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?
 * a. || 0.277 g || c. || 45.0 g ||
 * b. || 22.5 g || d. || 90.0 g ||

ANS: D The balanced chemical equation is.


 * || ** Feedback ** ||
 * ** A ** || Multiply the number of moles of water by the molar mass of water. ||
 * ** B ** || Multiply the number of moles by the mole ratio. ||
 * ** C ** || Balance the equation correctly. ||
 * ** D ** || Correct! ||

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Use the steps to solve stoichiometric problems. KEY: Stoichiometric mole-to-mass conversion

28. What is the mass of potassium chloride when 2.50 g of potassium reacts with excess of chlorine gas?
 * a. || 4.77 g || c. || 9.52 g ||
 * b. || 8.57 g || d. || 728 g ||

ANS: A The balanced chemical equation is.


 * || ** Feedback ** ||
 * ** A ** || Correct! ||
 * ** B ** || Calculate the mass of KCl using the molar mass as a conversion factor. ||
 * ** C ** || Balance the equation correctly. ||
 * ** D ** || Convert the grams of KCl to moles using the inverse of molar mass as the conversion factor. ||

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Use the steps to solve stoichiometric problems. KEY: Stoichiometric mass-to-mass conversion

29. How many moles of carbon dioxide are produced when 8.30 mol of ethanol reacts with excess of oxygen?
 * a. || 0.241 mol || c. || 16.6 mol ||
 * b. || 4.15 mol || d. || 24.9 mol ||

ANS: C The mole ratio of carbon dioxide to ethanol is 2:1.


 * || ** Feedback ** ||
 * ** A ** || Divide the unknown moles by the known moles. ||
 * ** B ** || Multiply the known number of moles by the mole ratio. ||
 * ** C ** || Correct! ||
 * ** D ** || Balance the equation correctly. ||

PTS: 1 **DIF: Bloom's Level 1|DOK 2** TOP: Use the steps to solve stoichiometric problems. KEY: Stoichiometric mole-to-mole conversion

30. Calculate the mass of citric acid when 2.60 mol of sucrose gas reacts with oxygen.
 * a. || 0.769 g || c. || 999 g ||
 * b. || 1.30 g || d. || 499 g ||

ANS: D The molar mass of citric acid is 192.044 g.


 * || ** Feedback ** ||
 * ** A ** || Divide the unknown moles by the known moles. ||
 * ** B ** || Multiply the known number of moles by the mole ratio and the molar mass. ||
 * ** C ** || Balance the equation correctly. ||
 * ** D ** || Correct! ||

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Use the steps to solve stoichiometric problems. KEY: Stoichiometric mole-to-mass conversion

31. Hydrofluoric acid reacts with 31.3 g of silica to produce hexafluorosilicic acid. Determine the percent yield of H2SiF6 if the actual yield is 60.3 g.
 * a. || 0.818% || c. || 31.8% ||
 * b. || 12.2% || d. || 81.8% ||

ANS: D Percent yield (actual yield/theoretical yield) × 100


 * || ** Feedback ** ||
 * ** A ** || Multiply the yield by 100 to calculate the percent yield. ||
 * ** B ** || Divide the actual yield by the theoretical yield. ||
 * ** C ** || The molar mass is incorrect. ||
 * ** D ** || Correct! ||

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Determine the percent yield for a chemical reaction. KEY: Percent yield

32. Copper reacts with 36.7 g of silver nitrate to produce copper(II) nitrate and silver. Determine the theoretical yield of Cu(NO3)2 if the actual yield is 31.29 g.
 * a. || 0.773 g || c. || 77.3 g ||
 * b. || 12.9 g || d. || 40.5 g ||

ANS: D Percent yield (actual yield/theoretical yield) × 100


 * || ** Feedback ** ||
 * ** A ** || Multiply the number of moles by the molar mass to obtain the theoretical yield. ||
 * ** B ** || The mole ratio is incorrect. ||
 * ** C ** || The molar mass is incorrect. ||
 * ** D ** || Correct! ||

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Determine the percent yield for a chemical reaction. KEY: Percent yield


 * COMPLETION **

1. When 200.2 g of sulfur reacts with 100.3 g of chlorine to produce disulfur dichloride, acts as the limiting reactant.

ANS: chlorine

PTS: 1 **DIF: Bloom's Level 2|DOK 1** TOP: Identify the limiting reactant in a chemical equation. KEY: Limiting reactant

2. Nitrogen acts as a(n) reactant in the production of ammonia.

ANS: excess

PTS: 1 **DIF: Bloom's Level 2|DOK 1** TOP: Identify the excess reactant and calculate the amount remaining after the reaction is complete. KEY: Excess reactant


 * SHORT ANSWER **

1. The mole ratios for the equation A + B → C + D are shown. Based on the mole ratios, write the balanced general chemical equation represented.
 * Substances || Mole Ratio || Substances || Mole Ratio ||
 * A:B || 2:3 || B:C || 3:1 ||
 * A:C || 2:1 || B:D || 1:2 ||
 * A:D || 1:3 || C:D || 1:6 ||

ANS: 2A + 3B → C + 6D

PTS: 1 **DIF: Bloom's Level 3|DOK 2**

2. Fill in the diagram with the conversion factors that would be used to determine the mass of Substance X required to form a certain mass of Substance Y.

ANS: A) molar mass of Substance X B) mole ratio between substance X and substance Y  C) molar mass of substance Y

PTS: 1 **DIF: Bloom's Level 2|DOK 2**

3. Fill in the boxes with the appropriate number of molecules of Substance Y and Substance Z to show the decomposition of Substance X in the correct ratios.

ANS: Six molecules of diatomic Substance Y, 4 molecules of Substance Z.

PTS: 1 **DIF: Bloom's Level 5|DOK 2**

4. A dimensional analysis setup to determine the mass of Fe that can be fully oxidized by 8.93 grams of O2 is shown. Explain what is incorrect in this setup and how to fix the setup to make it correct.

ANS: The mole ratio is reversed; it should be 4 moles of Fe/3 moles O2. This is because the number of moles of oxygen must be on the bottom of the fraction in order to cancel out the moles of oxygen in the previous fraction (molar mass of oxygen).

PTS: 1 **DIF: Bloom's Level 6|DOK 2**

5. Explain why an excess reagent is frequently used in chemical reactions.

ANS: An excess reagent will accomplish two things. It will ensure that the reaction goes to completion, since all of the limiting reagent will be able to be used. It will also help to make the reaction go faster.

PTS: 1 **DIF: Bloom's Level 2|DOK 1**

6. Find the mole ratio of lead (II) nitrate to potassium nitrate in this chemical reaction. KI + Pb(NO3)2 → PbI2 + KNO3

ANS: 1:2

PTS: 1 **DIF: Bloom's Level 3|DOK 2**

7. How many moles of zinc chloride can be produced from 15.3 moles of HCl? Zn + HCl → ZnCl2 + H2

ANS: 30.6 moles; HCl and ZnCl2 are in a 1:2 molar ratio.

PTS: 1 **DIF: Bloom's Level 3|DOK 2**

8. In this chemical reaction, 6 moles of each of the reactants is combined. Mg + N2 → Mg3N2 a) Which is the limiting reagent? Which is the excess reagent? b) How many moles of magnesium nitride can be made? c) How many moles of excess reagent will be left over?

ANS: a) Magnesium is limiting and nitrogen is excess; they react in a 3:1 molar ratio so there is proportionately less magnesium. b) 2 moles of product are formed; the mole ratio for the entire reaction is 3:1:1 c) Since 2 moles of nitrogen must be used, there will be 4 moles left over.

PTS: 1 **DIF: Bloom's Level 4|DOK 2**

9. How much chlorine should be produced if 84.2 grams of aluminum chloride and 68.4 grams bromine are combined? AlCl3 + Br2 → AlBr3 + Cl2

ANS: 30.3 grams The bromine is the limiting reagent; 0.428 moles of chlorine will be produced.

PTS: 1 **DIF: Bloom's Level 3|DOK 2**

10. 64.9 grams of potassium chloride are reacted with excess oxygen to produce potassium chlorate, according to the reaction:

KCl + O2 → KClO3

If 77.1 grams are produced, what is the percent yield of this reaction?

ANS: The theoretical yield would be 106.7 grams. Therefore, the percent yield is (77.1/106.7)x100, or 72.3%.

PTS: 1 **DIF: Bloom's Level 3|DOK 2**

11. Aluminum chloride and bromine react according to the reaction:

If the percent yield of chlorine is 83.1%, how much aluminum chloride must be used to guarantee a yield of 43.5 grams of chlorine?

ANS: With a percent yield of 83.1% and an actual yield of 43.5 grams, the theoretical yield must be at least (43.5 g)/(0.831) = 52.3 grams of chlorine gas. Using this value in stoichiometry, we find that 52.3 grams of chlorine requires 66.4 grams of aluminum chloride.

PTS: 1 **DIF: Bloom's Level 4|DOK 2**

12. What is stoichiometry?

ANS: Stoichiometry is a study of quantitative relationships between the amounts of reactants used and the products formed by a chemical reaction.

PTS: 1 **DIF: Bloom's Level 1|DOK 1** TOP: Identify the quantitative relationships in a balanced chemical equation. KEY: Stoichiometry

13. What is a mole ratio?

ANS: A mole ratio is the ratio between the numbers of moles of any two substances in a balanced chemical equation.

PTS: 1 **DIF: Bloom's Level 1|DOK 1** TOP: Determine the mole ratios from a balanced chemical equation. KEY: Mole ratio

14. Balance the following equation and determine the possible mole ratios.

ANS: The balanced chemical equation is:

The possible mole ratios are:

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Determine the mole ratios from a balanced chemical equation. KEY: Mole ratio

15. List the four steps to solve stoichiometric problems.

ANS: The four steps to solve stoichiometric problems are: a. Write a balanced chemical equation. b. Determine the moles of a given substance using a mass-to-mole conversion. c. Determine the moles of an unknown substance from the moles of the given substance. d. Determine the mass of an unknown substance from the moles of the unknown substance using a mole-to-mass conversion.

PTS: 1 **DIF: Bloom's Level 1|DOK 1** TOP: Explain the sequence of steps used in solving stoichiometric problems. KEY: Using stoichiometry


 * PROBLEM **

1. Titanium dioxide (TiO2) is an industrial chemical used as a white pigment in paint. The conversion of volatile TiCl4 to TiO2 occurs according to the reaction,. If 15.70 g of TiCl4 reacts in excess oxygen to form 10.40 g Cl2, what is the percent yield of the reaction?

ANS: 88.61%

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Determine the percent yield for a chemical reaction. KEY: Percent yield NOT: Percent yield = (actual yield/theoretical yield) * 100

2. Phosphorus pentachloride is formed when 17.2 g of chlorine gas react with 23.2 g of solid phosphorus (P2). Determine the reactant that is in excess.

ANS: Phosphorus is the reactant that is in excess.

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Identify the excess reactant and calculate the amount remaining after the reaction is complete. KEY: Excess reactant NOT: The actual ratio is less than the required ratio. The number of moles of chlorine needed in the reaction is 10, but only 1.29 moles of chlorine gas are available. Thus, chlorine acts as the limiting reactant and phosphorus as the excess reactant.

3. In a reaction, 82.00 g of sodium reacts with 74.00 g of ferric oxide to form sodium oxide and iron metal. Calculate the mass of solid iron produced.

ANS: 51.36 g

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Calculate the mass of a product when the amounts of more than one reactant are given. KEY: Product mass calculation NOT: First, calculate the actual ratio. Then, convert the number of moles of ferric oxide to the number of moles of iron.

4. In a reaction, 10.76 g of CaCO3, 10.51 g of HCl, and excess water produced 10.26 g of CaCl26H2O. Calculate the theoretical yield of calcium chloride hexahydrate.

ANS: 23.56 g

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Calculate the theoretical yield of a chemical reaction from data. KEY: Theoretical yield NOT: The theoretical yield is calculated by multiplying the number of moles of calcium chloride hexahydrate by the molar mass.

5. What is the percent yield for a reaction if the theoretical yield of C6H12 is 21 g and the actual yield recovered is only 3.8 g?

ANS: 18%

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Determine the percent yield for a chemical reaction. KEY: Percent yield NOT: Percent yield = (actual yield/theoretical yield) * 100


 * ESSAY **

1. A reaction is predicted to result in 75.0 grams of product being made, but only 58.3 grams are actually produced. Find the percent yield of this reaction, and explain several reasons why the reaction does not produce as much as predicted.

ANS: The percent yield is 77.7%. There may be procedural reasons why the yield is not 100%, such as a precipitate being left on filter paper or otherwise left behind. Liquids might stick to their containers or evaporate. Finally, there may be other reactions occurring at the same time, removing some of the reactants and preventing them from forming the desired product.

PTS: 1 **DIF: Bloom's Level 4|DOK 1**

2. Balance this equation. Then, show that the Law of Conservation of Matter is being obeyed at the particle level, the mole level, and the mass level.

Fe2O3 + C → Fe + CO

ANS:

Fe2O3 + 3C → 2Fe + 3CO

Particles: 2 atoms Fe, 3 atoms O, 3 atoms C as reactants; 2 atoms Fe, 3 atoms O, 3 atoms C as products Moles: 2 moles Fe, 3 moles O, 3 moles C as reactants; 2 moles Fe, 3 moles O, 3 moles Cas products Mass: 1(159.691) + 3(12.01) = 2 (55.847) + 3 (84.027); 195.721 g = 195.721

PTS: 1 **DIF: Bloom's Level 6|DOK 2**

3. Aluminum metal is burned in oxygen gas. a. Write the balanced equation for this reaction. b. If 5.433 g of aluminum is burned with 8.834 g of oxygen gas, what is the limiting reagent? c. What mass of product can be made?

ANS: a) 4Al(s) + 3O2(g) → 2Al2O3(s) b) The limiting reagent is the aluminum. c) 10.3 grams of aluminum oxide can be produced.

PTS: 1 **DIF: Bloom's Level 4|DOK 2**

4. You have 83.6 grams of H2 and 257 grams of N2 which combine according to the following equation:

N2 + 3H2 → 2NH3

Which reactant is the limiting reagent? Explain how you can tell.

ANS:

The molar amounts in this reaction are 9.17 moles nitrogen and 41.5 moles hydrogen, based on the molar masses of each compound. The mole ratio for the entire reaction is 1:3:2. Since 41.5 moles is more than 3 times larger than 9.17 moles, the nitrogen is the limiting reagent. It will only react with 27.5 moles of hydrogen, leaving 14 moles of hydrogen in excess.

PTS: 1 **DIF: Bloom's Level 4|DOK 2**

5. 8.344 grams of sodium hydroxide is reacted with 14.290 grams of magnesium nitrate, according to the reaction NaOH + Mg(NO3)2 → NaNO3 + Mg(OH)2. a) Write the balanced chemical equation for this reaction. b) Identify the limiting reagent. c) Determine the theoretical yield of magnesium hydroxide.

ANS:

a) 2 NaOH + Mg(NO3)2 → 2 NaNO3 + Mg(OH)2. b) The limiting reagent is NaOH c) There are 0.1043 moles, or 6.083 grams, of magnesium hydroxide.

PTS: 1 **DIF: Bloom's Level 5|DOK 2**

6. Represent the reaction between nitrogen and hydrogen in terms of: a. Particles b. Moles c. Mass

ANS: a. 1 molecule N2 + 3 molecules H2 → 2 molecules NH3 b. 1 mole N2 + 3 moles H2 → 2 moles NH3 c. 28.02 g N2 + 6.06 g H2 → 34.08 g NH3

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Determine the mole ratios from a balanced chemical equation. KEY: Mole ratio

7. Represent the reaction between zinc and nitric acid in terms of: a. Particles b. Moles c. Mass

ANS: a. 4 atoms Zn + 10 molecules HNO3 → 4 formula units Zn(NO3)2 + 1 molecule N2O + 5 molecules H2O b. 4 moles Zn + 10 moles HNO3 → 4 moles Zn(NO3)2 + 1 mole N2O + 5 moles H2O c. 261.56 g Zn + 630.2 g HNO3 → 757.56 g Zn(NO3)2 + 44.02 g N2O + 90.10 g H2O

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Determine the mole ratios from a balanced chemical equation. KEY: Mole ratio

8. In the equation for the combustion of butane, show that the law of conservation of mass is observed. Interpret the equation for the combustion of butane in terms of: a. Representative particles b. Moles c. Mass

ANS: The equation for the combustion of butane is. a. 2 molecules C4H10 + 13 molecules O2 → 8 molecules CO2 + 10 molecules H2O. b. 2 moles C4H10 + 13 moles O2 → 8 moles CO2 + 10 moles H2O. c. 116 g C4H10 + 416 g O2 → 352 g CO2 + 180 g H2O. Mass of reactants Mass of products 532 g.

PTS: 1 **DIF: Bloom's Level 2|DOK 2** TOP: Determine the mole ratios from a balanced chemical equation. KEY: Mole ratio

9. A 45.00-g sample of silver nitrate is mixed with 55.00 g of hydrochloric acid to form a white precipitate of silver chloride. After the solution is filtered and dried, a white precipitate of mass 33.50 g is collected. a. Determine the limiting reactant. b. Determine the theoretical yield of silver chloride. c. Determine the percent yield of silver chloride.

ANS: a. Silver nitrate is the limiting reactant. b. The theoretical yield of silver chloride is 38.02 g. c. The percent yield of silver chloride is 88.09%.

PTS: 1 **DIF: Bloom's Level 3|DOK 2** TOP: Determine the percent yield for a chemical reaction. KEY: Percent yield

7. Given the reaction: N2 (g) + 3 H2 (g) -> 2 NH3 (g) What is the mole-to-mole ratio between nitrogen gas and hydrogen gas? a. 1:2 b. 1:3 c. 2:2 d. 2:3

ANS: (b)

PTS: 1 **DIF**: **Bloom's Level 3|DOK 2**