Download Chemistry I/IH Chapter 9 Stoichiometry PRACTICE TEST Multiple

Chemistry I/IH Chapter 9 Stoichiometry PRACTICE TEST
Multiple Choice Identify the choice that best completes the statement or answers the question.
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1. The coefficients in a chemical equation represent the
a. masses, in grams, of all reactants and products.
b. relative numbers of moles of reactants and products.
c. number of atoms in each compound in a reaction.
d. number of valence electrons involved in the reaction.
2. A balanced chemical equation allows one to determine the
a. mole ratio of any two substances in the reaction.
b. energy released in the reaction.
c. electron configuration of all elements in the reaction.
d. mechanism involved in the reaction.
3. To balance a chemical equation, it may be necessary to adjust the
a. coefficients.
b. subscripts.
c. formulas
d. charges on ions
4. Given the equation
, the starting mass of A, and its molar mass, and you are asked to determine
the moles of C produced, your first step in solving the problem is the multiply the given mass of A by
a.
b.
c.
d.
5. In the chemical reaction represented by the equation wA + xB  yC + zD, a comparison of the number of moles
of A to the number of moles of C would be a(n)
a. mass ratio.
b. mole ratio.
c. electron ratio.
d. energy ratio.
6. In the reaction represented by the equation N2 + 3H2  2NH3, what is the mole ratio of nitrogen to ammonia?
a. 1:1
b. 1:2
c. 1:3
d. 2:3
Short Answer: Show all your work. Circle your answers on the far right of the page. Don’t forget units&sig figs.
7. Explain why it would be rare to have two limiting reactants in the same chemical reaction.
8. Determine the number of moles of H2O that are produced when 17 mol of H2 are consumed using the equation
2H2 + O2 2H2O.
Problem
9. What mass is grams of potassium chlorate is consumed when 212 moles of potassium chloride are produced
according to the following equation?
10. How many moles of Ag can be produced if 116 g of Cu are reacted with excess AgNO3 according to the equation
Cu(s) + 2AgNO3(aq)  2Ag(s) + Cu(NO3)2(aq)?
11. What mass of PCl3 forms in the reaction of 27.0 g P4 with 95.2 g Cl2 ?
12.
Use the equation 2Al + 3Cl2
2 AlCl3 to answer the following questions.
a. Determine the number of grams of product that can be produced from 28.5 grams Al and 41.9 grams Cl2 .
b. (EXTRA CREDIT)Calculate the grams of excess reactant remaining when the reaction is complet
Chemistry I/IH Chapter 9 Stoichiometry PRACTICE TEST
KEY
MULTIPLE CHOICE
1. ANS: B
3. ANS: A
5. ANS: B
2. ANS: A
4. ANS: C
6. ANS: B
SHORT ANSWER
7. ANS: Reactions continue until all of one of the reactants is used up. The chemical that runs out is called
the limiting reactant. It would be hard to measure chemicals exactly so that both (or all) of the chemicals
run out at the same time.
8. ANS: 17 mol H2 x 2 mol H2O = 17 mol
2 mol H2
PROBLEM
9.
10.
11. Assuming that P4 is the limiting reagent:
Assuming that Cl2 is the limiting reagent:
Since the smaller amount of product is formed from P4, it is the limiting reagent. The mass of product formed
is: 120 g PCl3
12. a. Find the limiting reagent first. Assume that the other reactant is in excess when you calculate the moles of
product formed from the first reactant. The reactant that gives the smaller amount of product is limiting. It
gives the maximum amount of product for the reaction. Set up the conversions and calculate.
28.5g Al x 1 mol Al
x 2 mol AlCl3 x 133.33g AlCl3 = 141 g AlCl3
26.98 g Al
2 mol Al
1 mol AlCl3
41.9 g Cl2 x 1 mol Cl2
70.9 g Cl2
x 2mol AlCl3
3 mol Cl2
x 133.33g AlCl3 = 52.5 g AlCl3
1 mol AlCl3
(This means that chlorine is the limiting reactant & aluminum is the excess reactant.)
b. Amount of excess reactant you start with – amount excess reactant used = amount remaining (in excess)
28.5 g Al - ??? = amount in excess
b1. Using the amount of product formed as the given, (calculated in the prior calculations), solve for the
amount of excess reactant USED.
52.5 g AlCl3 x 1 mol AlCl3 x
2 mol Al
x 26.98 g Al = 10.6 g Al
133.33 g AlCl3
2 mol AlCl3
1 mol Al
b2. Now, subtract the amount of excess reactant USED from the amount of excess reactant you started with.
This is the amount of excess reactant left over!!!
28.5g Al – 10.6 g Al = 17.9 g Al in excess (left over/remaining)