Lesson C08 Solutions
78. Define the problem: Given the volume and molarity of a solution containing one reactant, the balanced
chemical equation, and a series of steps describing the calculation, determine which of the steps is not
correct, and correctly determine the mass of the other reactant in a specific volume of its solution.
Develop a plan: Check each step to see if it is right or wrong. If it is wrong, correct it.
Execute the plan:
(a) Step (i) is correct.
6.42 mL ×
1L
9.580 × 10−2 mol NaOH
×
= 6.15 ×10−4 mol NaOH
1000 mL
1L
Step (ii) is incorrect.
The equation given in the Question is not balanced. The balanced equation is:
C3 H5 O(COOH)3 (aq) + 3 NaOH(aq)
Na 3 C3 H5 O(COO)3 (aq) + 3 H2 O(l)
So, the equation stoichiometry gives 1 mol citric acid reacting with 3 mol of NaOH.
6.15 ×10−4 mol NaOH ×
1 mol citric acid
= 2.05 × 10−4 mol citric acid
3 mol NaOH
Step (ii) has a 3 in front of the moles of citric acid and a 1 in front of the NaOH, resulting in
multiplication by 3 instead of division by 3, so the moles of citric acid calculated there are wrong.
Step (iii) is incorrect, because it uses the erroneous answer from Step (ii); however, the
calculation it shows is correct:
2.05 ×10−4 mol citric acid ×
192 .12 g citric acid
= 0.0394 g citric acid
1 mol citric acid
Step (iv) is incorrect, because it uses a different answer than the correct one from Step (iii)
and the significant figures on the volume are incorrect; however, the calculation it shows is
correct:
0.0394 g citric acid
= 3.94 × 10 −3 g citric acid in 1 mL of soft drink
10.0 mL
–3
(b) The right answer is 3.94 × 10 g citric acid in 1 mL of soft drink.
Check your answer: The moles of citric acid that react must be less than the moles of NaOH that react.
This answer now makes sense.
79. Define the problem: Given the volume and molarity of a solution containing one reactant, the balanced
chemical equation, and a series of steps describing the calculation, determine which of the steps is not
correct, and correctly determine the mass of the other reactant in a specific volume of its solution.
Develop a plan: Check each step to see if it is right or wrong. If it is wrong, correct it.
Execute the plan:
(a) Step (i) is incorrect.
27 .85 mL ×
1L
0.102 mol Br 2
×
= 2.84 ×10 −3 mol Br2
1000 mL
1L
Step (ii) is incorrect, because it uses the erroneous answer from Step (i); however, the
calculation it shows is correct:
The balanced equation is
C6 H8 O6 (aq) + Br2 (aq)
2 HBr(aq) + C6 H6 O6 (aq)
So, the equation stoichiometry gives 1 mol C6 H8 O6 reacting with 1 mol of Br2 .
2.84 × 10
−3
mol Br2 ×
1 mol C6 H 8O6
−3
= 2.84 × 10 mol C 6H 8 O6
1 mol Br2
Step (iii) is incorrect, because it uses the erroneous answer from Step (ii); however, the
calculation it shows is correct (except we try to use molar masses with at least one more
significant figure than the rest of the data to help prevent round off errors):
2.84 × 10
−3
mol C6 H 8O 6 ×
176 .1238 g C6 H 8O6
= 0.500 g C 6H 8 O6
1 mol C 6H 8 O6
Step (iv) is incorrect, because it isn’t needed. Furthermore, it uses the incorrect answer from Step
(iii), the significant figures on the mass in the denominator are incorrect, and the calculation it
shows is als o not completely correct:
0.500 g C6 H 8O 6
C H O
= 0.500 6 8 6 = fraction of the tablet that is C6 H8 O6
1.00 g tablet
tablet
To get the mass of C6 H8 O6 in the tablet, we need to multiply the mass of the tablet by the fraction
– which is essentially undoing the calculation in Step (iv). This step was unnecessary, since the
calculation at the end of (iii) gives the mass of C6 H8 O6 in the tablet.
(b) The result of Step (iii) shows that there are 0.500 g C6 H8 O6 in the tablet.
Check your answer: The mass of vitamin C is now smaller than the mass of the tablet! This answer
now makes sense.
80. Define the problem: Given the mass of one reactant, the balanced chemical equation for a reaction, and
the volume of a solution containing the other reactant needed for a complete reaction, determine the
molarity of the second solution.
Develop a plan: The mass and molar mass can be used to find the moles of one reactant. Then we will
use the equation stoichiometry to find out moles of the other reactant needed. Then we will use the
moles and volume in liters to determine the molarity. Note: It is NOT appropriate to use the dilution
equation when working with reactions!
Execute the plan:
We learn from the balanced equation that 1 mol of Na 2 CO3 reacts with 2 mol HCl.
2.050 g Na 2 CO 3 ×
1 mol Na 2CO 3
2 mol HCl
−2
×
= 3.868 × 10 mol HCl
105.9885 g Na 2CO3 1 mol Na 2 CO3
32 .45 mL HCl solution×
1L
= 0.03245 L HCl solution
1000 mL
3.868 × 10−2 mol HCl
= 1.192 M HCl solution
0.03245 L HCl solution
The three separate calculations above can be consolidated into one calculation as follows:
2.050 g Na 2CO 3
1 mol Na 2CO3
2 mol HCl
1000 mL
×
×
×
= 1.192 M HCl solution
32.45 mL HCl solution 105.9885 g Na 2CO 3 1 mol Na 2CO 3
1L
Both methods will give the right answer, however the consolidated calculation eliminates the need to
write down unnecessary intermediate answers and helps eliminate round-off errors.
Check your answer: The units cancel appropriately, and the moles and liters are nearly the same value
so it makes sense that the molarity is near 1.
81. Define the problem: Given the mass of one reactant, the balanced chemical equation for a reaction, and
the volume of a solution containing the other reactant needed for a complete reaction, determine the
molarity of the second solution.
Develop a plan: Follow the method described in the answer to Question 80
–
Execute the plan: We learn from the balanced equation that 1 mol of HC8 H4 O4 reacts with 1 mol
–
OH .
–
We learn from the formulas that 1 mol of HC8 H4 O4 is in 1 mol of KHC8 H4 O4 (also called KHP) and
–
1 mol OH is in 1 mol of NaOH.
0.902 g KHP
1 mol KHP
1 mol HC 8H 4 O−4
1 mol OH−
×
×
×
−
26.45 mL NaOH solution 204.2210 g KHP
1 mol KHP
1 mol HC 8 H4 O4
×
1 mol NaOH
1 mol OH
−
×
1000 mL
= 0.167 M NaOH solution
1L
Check your answer: The units cancel appropriately. The concentration of the NaOH solution is a
reasonable size.
82. Define the problem: Given the volume and concentration of a solution containing one reactant, the
balanced chemical equation for a reaction, and the mass of an impure sample of a salt of the second
reactant, determine the percent purity of the impure sample.
Develop a plan: The volume and molarity of the first reactant are used to calculate the moles.
Equation and formula stoichiometry can be used to find the moles of the salt. Then we will use the
molar mass of the salt to find the mass of the salt. Comparing this mass to the mass of the impure
sample we can determine the percent purity. Note: It is NOT appropriate to use the dilution equation
when working with reactions!
2–
Execute the plan: We learn from the balanced equation that 2 mol of S2 O3 reacts with 1 mol I2 .
2–
Formula stoichiometry tells us that 1 mol of S2 O3 comes from 1 mol of Na 2 S2 O3 .
1L
0.246 mol I 2
−3
40.21 mL I 2 solution×
×
= 9.89 ×10 mol I 2
1000 mL 1 L I 2 solution
9.89 × 10
−3
mol I 2 ×
2 mol S 2O2−
3 × 1 mol Na 2S 2O3 × 158.1098 g Na 2S 2 O3 = 3.13 g Na S O
2 2 3
2−
1 mol I 2
1 mol Na 2S 2 O3
1 molS 2 O3
The percent purity is calculated by dividing the mass of Na 2 S2 O3 by the mass of the impure sample and
multiplying by 100 %.
3.13 g Na 2S 2O 3
× 100 % = 96 .8 % pure
3.232 g impure sample
Check your answer: The units cancel appropriately, and the two masses are very similar, so it makes
sense that the percentage is nearly 100 %.
83. Define the problem: Given the volume and concentration of a solution containing one reactant, the
balanced chemical equation for a reaction, and the mass of a mixture containing two salts – one of
which is the salt of the second reactant, determine the weight percent of the reactive salt in the mixture.
Develop a plan: The volume and molarity of the first reactant are used to calculate the moles.
Equation stoichiometry can be used to find the moles of the reactive compound. Then we will use the
molar mass of the reactive compound to find the mass. Then we will compare that to the mass of the
mixture to determine the weight percent. Note: It is NOT appropriate to use the dilution equation
when working with reactions!
Execute the plan: We learn from the balanced equation that 1 mol of H2 C2 O4 reacts with 2 mol
NaOH.
29 .58 mL NaOH solution×
1L
0.550 mol NaOH
×
1000 mL 1 L NaOH solution
1 mol H 2C 2O 4 90.0348 g H2 C2O4
×
×
= 0.732 g H2 C2 O4
2 mol NaOH
1 mol H 2C 2O 4
The weight percent is calculated by dividing the mass of H2 C2 O4 by the mass of the mixture and
multiplying by 100 %.
0.732 g H 2C 2O 4
× 100 % = 16 .1 % H 2 C2O 4
4.554 g mixture
Check your answer: The units cancel appropriately, and the mass of H2 2 C2 O4 is significantly smaller
than the mass of the sample, so it makes sense that the percentage is so low.
110. Define the problem: Given an unknown diprotic acid, the volume and concentration of a solution
containing the base that is used to neutralize it, the balanced chemical equation for the neutralization
reaction, determine the molar mass of the acid.
Develop a plan: The volume and molarity of the first reactant are used to calculate the moles.
Equation stoichiometry can be used to find the moles of the acid. Then we will divide the mass by
the moles to get molar mass. Note: It is NOT appropriate to use the dilution equation when working
with reactions!
Execute the plan: We learn from the balanced equation that 1 mol of H2 A reacts with 2 mol NaOH.
36 .04 mL NaOH solution×
1L
0.509 mol NaOH
1 mol H 2 A
×
×
= 0.00917 mol H2 A
1000 mL 1 L NaOH solution 2 mol NaOH
0.954 g H 2 A
g H2 A
= 104
0.00917 mol H 2 A
mol H2 A
The two separate calculations above can be consolidated into one calculation as follows:
0.956 g H 2 A
1000 mL 1 L NaOH solution 2 mol NaOH
×
×
×
= 104
36.04 mL NaOH solution
1L
0.509 mol NaOH
1 mol H 2 A
g H2 A
mol H 2 A
Both methods will give the right answer, however the consolidated calculation eliminates the need to
write down unnecessary intermediate answers and helps eliminate round-off errors.
Check your answer: The units cancel appropriately, and the moles are nearly 0.01 times the mass
value so it makes sense that the molar mass is near 100.