11.1.3: Stoichiometry and the Ideal Gas Law
- Page ID
- 409042
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Convert from gas information of one substance to gas information of another substance in a chemical reaction.
Gas Stoichiometry Conversions
The third pathway we will look at is starting with gas information of one chemical in an equation and ending with gas information of another. See the highlighted portion below.
Here's an example of how it will work.
Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give \(\ce{SO2}\), followed by the reaction of \(\ce{SO2}\) with \(\ce{O2}\) in the presence of a catalyst to give \(\ce{SO3}\), which reacts with water to give \(\ce{H2SO4}\). The overall chemical equation is as follows:
\[\ce {2S(s) + 3O2(g) + 2H2O(l) \rightarrow 2H2SO4(aq)} \nonumber \]
What volume of O2 (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H2SO4?
Given: reaction, temperature, pressure, and mass of one product
Asked for: volume of gaseous reactant
Strategy:
A Calculate the number of moles of H2SO4 in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of \(\ce{O2}\) required.
B Use the ideal gas law to determine the volume of \(\ce{O2}\) required under the given conditions. Be sure that all quantities are expressed in the appropriate units.
Solution:
mass of \(\ce{H2SO4}\) → moles \(\ce{H2SO4}\) → moles \(\ce{O2}\) → liters \(\ce{O2}\)
A We begin by calculating the number of moles of H2SO4 in 1.00 ton:
\[\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4 \nonumber\]
We next calculate the number of moles of \(\ce{O2}\) required:
\[\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2 \nonumber\]
B After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O2:
\[\begin{align*} V&=\dfrac{nRT}{P} \\[4pt] &=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}} \\[4pt] &=3.43\times10^5\;L \end{align*}\]
The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry.
If you would like some additional practice applying your mole map to ideal gas stoichiometry, there are additional examples here and here. Please note that all of these examples have a similar form and that not all ideal gas stoichiometry problems will follow this particular form. However, it is probably useful to master this form prior to some of the additional considerations which follow.
Charles used a balloon containing approximately 31,150 L of \(\ce{H2}\) for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation:
\[\ce{ Fe(s) + 2 HCl(aq) \rightarrow H2(g) + FeCl2(aq)} \nonumber\]
How much iron (in kilograms) was needed to produce this volume of \(\ce{H2}\) if the temperature were 30°C and the atmospheric pressure was 745 mmHg?
- Answer
-
68.6 kg of Fe (approximately 150 lb)
Additional applications of relating the mole concept to the ideal gas law can be found here, including a discussion of STP (the conditions necessary to use the original mole map.)