# Practice Problems (Chapter 5): Stoichiometry

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﻿Practice Problems (Chapter 5): Stoichiometry

Part I: Using the conversion factors in your tool box

Tool Box: To convert between

g A mol A mol A particles A

mol A mol B

Use

molar ratio

From

periodic table memory

coeff. in bal. eqn.

KEY

CHEM 30A

1. How many moles CH3OH are in 14.8 g CH3OH?

1 mol CH3OH

14.8 g CH3OH

= 0.462 mol CH3OH

32.042 g CH3OH

CH3OH = 1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol) = 32.042 g/mol

2. What is the mass in grams of 1.5 x 1016 atoms S?

1 mol S 1.5 x 1016 atoms S 6.022 x 1023 atoms S

32.06 g S 1 mol S

= 8.0 x 10?7 g S

3. How many molecules of CO2 are in 12.0 g CO2?

1 mol CO2 12.0 g CO2

44.01 g CO2

6.022 x 1023 molecules CO2 = 1.64 x 1023 molecules CO2 1 mol CO2

CO2 = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol

4. What is the mass in grams of 1 atom of Au?

1 mol Au 1 atom Au 6.022 x 1023 atoms Au

197.0 g Au = 3.271 x 10?22 g Au

1 mol Au

Part II: Stoichiometry problems

5. If 54.7 grams of propane (C3H8) and 89.6 grams of oxygen (O2) are available in the balanced

combustion reaction to the right:

a) Determine which reactant is the limiting reactant. b) Calculate the theoretical yield of CO2 in grams.

C3H8 + 5 O2 3 CO2 + 4 H2O

54.7 g 89.6 g

Way #1:

54.7 g C3H8

1 mol C3H8 44.094 g C3H8

3 mol CO2 1 mol C3H8

44.01 g CO2 = 163.787 g CO2 1 mol CO2

89.6 g O2

L.R.

1 mol O2 32.00 g O2

Way #2:

54.7 g C3H8

1 mol C3H8 44.094 g C3H8

89.6 g O2

L.R.

(actually have)

1 mol O2 32.00 g O2

3 mol CO2 5 mol O2

5 mol O2 1 mol C3H8 3 mol CO2 5 mol O2

44.01 g CO2 1 mol CO2

= 73.937 g CO2 forms less product

theoretical yield CO2

32.00 g O2 1 mol O2 44.01 g CO2 1 mol CO2

= 198.485 g O2 needed only have 89.6 g O2

= 73.937 g CO2

theoretical yield CO2

Limiting Reactant: _______O__2 _______ Theoretical Yield: ____7_3_.9__g_C__O_2____

6. A reaction has a theoretical yield of 124.3 g SF6, but only 113.7 g SF6 are obtained in the lab, what is

the percent yield of SF6 for this reaction?

% yield SF6 =

actual yield SF6

113.7 g SF6

theoretical yield SF6 (100%) = 124.3 g SF6 (100%) = 91.47224457 % yield SF6

7. If 23.2 grams of butane (C4H10) and 93.7 grams of oxygen (O2) are available in the following

reaction:

_2___ C4H10 + _1_3__ O2 __8__ CO2 + __1_0_ H2O

23.2 g

93.7 g

Check:

C 8 H 20 O 26 charge 0

a) Balance the equation for the reaction. b) Determine which reactant is the limiting reactant. c) Calculate the theoretical yield of CO2 in grams.

Way #1:

23.2 g C4H10

L.R.

1 mol C4H10 58.120 g C4H10

8 mol CO2 2 mol C4H10

44.01 g CO2 1 mol CO2

theoretical yield CO2

= 70.271 g CO2 forms less product

93.7 g O2

Way #2:

23.2 g C4H10

1 mol O2 32.00 g O2

1 mol C4H10 58.120 g C4H10

23.2 g C4H10

L.R.

1 mol C4H10 58.120 g C4H10

(actually have)

8 mol CO2 13 mol O2

13 mol O2 2 mol C4H10 8 mol CO2 2 mol C4H10

44.01 g CO2 = 79.303 g CO2 1 mol CO2

32.00 g O2 1 mol O2

44.01 g CO2 1 mol CO2

= 83.028 g O2 only need have 93.7 g O2

= 70.271 g CO2

theoretical yield CO2

Limiting Reactant: _____C__4H__1_0 ______ Theoretical Yield: ___7_0_._3_g__C_O__2____

d) If the actual yield of CO2 is 69.2 g CO2, what is the percent yield?

actual yield CO2

69.2 g CO2

% yield CO2 = theoretical yield CO2 (100%) = 70.271 g CO2 (100%) = 98.476% yield CO2