Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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PRACTICE EXAMPLE A: What volume of the aluminum-copper alloy described in Example 4-5 must be dissolved in an excess of HCl(aq) to produce 1.00 g H₂? [Hint: Think of this as the "inverse" of Example 4-5.]
PRACTICE EXAMPLE B: A fresh sample of the aluminum-copper alloy described in Example 4-5 yielded 1.31 g H₂. How many grams of copper were present in the sample?
![EXAMPLE 4-5
Additional Conversion Factors in a Stoichiometric Calculation:
Volume, Density, and Percent Composition
d
An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of
L8/cm.A 0.691 cm³ piece of the alloy reacts with an excess of HCI(aq). If we assume that all the Al but none
of the Cu reacts with HCI(aq), what is the mass of H, obtained? Refer to reaction (4.2).
Analyze
A simple approach to this calculation is outlined below, Each numbered arrow refers to a conversion factor that
changes the unit on the left to the one on the right.
cm alloy
1
g alloy
2.
g Al mol Al → mol H,
3
g H2
The calculation can be done in five distinct steps, or with a single setup in which the five conversions are
performed in sequence.
Solve
Using a stepwise approach, we proceed as follows.
Convert from_volume of alloy to
grams of alloy by using the density as
a conversion factor.
2.85 g alloy
?g alloy = 0.691 cm³ alloy ×
1 cm alloy
= 1.97 g alloy
Convert from grams of alloy to grams
of Al by using the percentage by
mass of Al as a conversion factor.
93.7 g Al
?g Al
1.97 g alloy X
100 g alloy
= 1.85 g Al
Convert from grams of Al to moles
of Al by using the molar mass of Al as
a conversion factor.
1 mol Al
? mol Al = 1.85 g Al X
26.98 g Al
0.0684 mol Al
%3D
Convert from moles of Al to moles of
3 mol H2
H2 by using the stoichiometric factor.
? mol H2 = 0.0684 mol Al ×
2 mol Al
= 0.103 mol H2
Convert from moles of H2 to grams of
H2 using the molar mass of H2 as a
conversion factor.
2.016 g H2
1 mol H2
? g H2 = 0.103 mol H2 ×
= 0.207 g H2
Remember to store intermediate results without rounding off. When all of the steps are combined into a single
calculation, we do not have to write down intermediate results and we reduce the likelihood of rounding errors.
93.7 g Al
2.016 g H2
1 mol H2
2.85 g alloy
1 mol Al
3 mol H2
? g H2 = 0.691 cm alloy ×
1 cm alloy
100 g alloy 26.98 g Al
2 mol Al
= 0.207 g H2
Assess
The units work out properly, but we must evaluate whether the answer is a reasonable one. The molar masses
of Al and H2 are approximately 27 g/mol and 2 g/mol, respectively. Equation (4.2) tells us that 1 mole of Al,
which weighs approximately 27 g, produces 1.5 mol of H2, which weighs 1.5 x 2 = 3 g. Thus, 27 g Al produces
approximately 3 g H2 and, thus, 2.7 g Al produces approximately 0.3 g H2. In this example, we are dealing with
less than 2.7 g of Al; therefore, we expect less than 0.3 g of H2. The answer, 0.207 g H2, is reasonable.
PRACTICE EXAMPLE A: What volume of the aluminum-copper alloy described in Example 4-5 must be dissolved
in an excess of HCl(aq) to produce 1.00 g H2? [Hint: Think of this as the "inverse" of Example 4-5.]
PRACTICE EXAMPLE B: A fresh sample of the aluminum-copper alloy described in Example 4-5 yielded 1.31 g H2.
How many grams of copper were present in the sample?](https://content.bartleby.com/qna-images/question/ae597109-beaa-4931-b9c6-e5443667ecac/fbee402b-4796-480e-9e0b-49cb49c469c0/a8h4rk_processed.jpeg)
Transcribed Image Text:EXAMPLE 4-5
Additional Conversion Factors in a Stoichiometric Calculation:
Volume, Density, and Percent Composition
d
An alloy used in aircraft structures consists of 93.7% Al and 6.3% Cu by mass. The alloy has a density of
L8/cm.A 0.691 cm³ piece of the alloy reacts with an excess of HCI(aq). If we assume that all the Al but none
of the Cu reacts with HCI(aq), what is the mass of H, obtained? Refer to reaction (4.2).
Analyze
A simple approach to this calculation is outlined below, Each numbered arrow refers to a conversion factor that
changes the unit on the left to the one on the right.
cm alloy
1
g alloy
2.
g Al mol Al → mol H,
3
g H2
The calculation can be done in five distinct steps, or with a single setup in which the five conversions are
performed in sequence.
Solve
Using a stepwise approach, we proceed as follows.
Convert from_volume of alloy to
grams of alloy by using the density as
a conversion factor.
2.85 g alloy
?g alloy = 0.691 cm³ alloy ×
1 cm alloy
= 1.97 g alloy
Convert from grams of alloy to grams
of Al by using the percentage by
mass of Al as a conversion factor.
93.7 g Al
?g Al
1.97 g alloy X
100 g alloy
= 1.85 g Al
Convert from grams of Al to moles
of Al by using the molar mass of Al as
a conversion factor.
1 mol Al
? mol Al = 1.85 g Al X
26.98 g Al
0.0684 mol Al
%3D
Convert from moles of Al to moles of
3 mol H2
H2 by using the stoichiometric factor.
? mol H2 = 0.0684 mol Al ×
2 mol Al
= 0.103 mol H2
Convert from moles of H2 to grams of
H2 using the molar mass of H2 as a
conversion factor.
2.016 g H2
1 mol H2
? g H2 = 0.103 mol H2 ×
= 0.207 g H2
Remember to store intermediate results without rounding off. When all of the steps are combined into a single
calculation, we do not have to write down intermediate results and we reduce the likelihood of rounding errors.
93.7 g Al
2.016 g H2
1 mol H2
2.85 g alloy
1 mol Al
3 mol H2
? g H2 = 0.691 cm alloy ×
1 cm alloy
100 g alloy 26.98 g Al
2 mol Al
= 0.207 g H2
Assess
The units work out properly, but we must evaluate whether the answer is a reasonable one. The molar masses
of Al and H2 are approximately 27 g/mol and 2 g/mol, respectively. Equation (4.2) tells us that 1 mole of Al,
which weighs approximately 27 g, produces 1.5 mol of H2, which weighs 1.5 x 2 = 3 g. Thus, 27 g Al produces
approximately 3 g H2 and, thus, 2.7 g Al produces approximately 0.3 g H2. In this example, we are dealing with
less than 2.7 g of Al; therefore, we expect less than 0.3 g of H2. The answer, 0.207 g H2, is reasonable.
PRACTICE EXAMPLE A: What volume of the aluminum-copper alloy described in Example 4-5 must be dissolved
in an excess of HCl(aq) to produce 1.00 g H2? [Hint: Think of this as the "inverse" of Example 4-5.]
PRACTICE EXAMPLE B: A fresh sample of the aluminum-copper alloy described in Example 4-5 yielded 1.31 g H2.
How many grams of copper were present in the sample?
![PRACTICE EXAMPLE A: What volume of the aluminum-copper alloy described in Example 4-5 must be dissolved
in an excess of HCl(aq) to produce 1.00 g H2? [Hint: Think of this as the "inverse" of Example 4-5.]
PRACTICE EXAMPLE B: A fresh sample of the aluminum-copper alloy described in Example 4-5 yielded 1.31 g H2.
How many grams of copper were present in the sample?](https://content.bartleby.com/qna-images/question/ae597109-beaa-4931-b9c6-e5443667ecac/fbee402b-4796-480e-9e0b-49cb49c469c0/et9rksg_processed.jpeg)
Transcribed Image Text:PRACTICE EXAMPLE A: What volume of the aluminum-copper alloy described in Example 4-5 must be dissolved
in an excess of HCl(aq) to produce 1.00 g H2? [Hint: Think of this as the "inverse" of Example 4-5.]
PRACTICE EXAMPLE B: A fresh sample of the aluminum-copper alloy described in Example 4-5 yielded 1.31 g H2.
How many grams of copper were present in the sample?
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