8.6: Stoichiometry and the Ideal Gas Law
- Page ID
- 372182
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- To relate the amount of gas consumed or released in a chemical reaction to the stoichiometry of the reaction.
- To understand how the ideal gas equation and the stoichiometry of a reaction can be used to calculate the volume of gas produced or consumed in a reaction.
You may remember we had previously discussed the concept of a mole roadmap. This was done because the mole concept we had introduced earlier can be connected to many other concepts that we had not introduced yet. Now we can make an additional connection for our mole roadmap: gases. You may note on the example mole road map gases were already included. That existing relationship only works some of the time: when the gas is at a specific temperature and pressure. Now we have learned how to find the moles of a gas using the ideal gas law. We can use the ideal gas law to relate moles to any other of the gas properties, not just volume of a gas at a specified temperature and pressure.
Take a moment to revise your existing mole roadmap to incorporate properties of gases. As you had done previously with stoichiometry, you should be able to convert from any point on the roadmap to any other point. One important thing to remember here is that you can only use the ideal gas law when you have moles of a gas. If you have a substance in any other physical state, you will likely need to use your mole roadmap to convert to moles of a gas prior to using the ideal gas law. In some cases, you may start with gas properties and need to determine the amount of some other substance which is not a gas. Your mole roadmap should be created such that it can solve this variety of problems. Once you have made some changes to your mole roadmap, see if you could use it to answer the following example problems.
What volume of carbon dioxide gas is produced at STP by the decomposition of 0.150 g \(\ce{CaCO_3}\) via the equation:
\[ \ce{CaCO3(s) \rightarrow CaO(s) + CO2(g)} \nonumber\]
Solution
Begin by converting the mass of calcium carbonate to moles.
\[ \dfrac{0.150\;g}{100.1\;g/mol} = 0.00150\; mol \nonumber\]
The stoichiometry of the reaction dictates that the number of moles \(\ce{CaCO_3}\) decomposed equals the number of moles \(\ce{CO2}\) produced. Use the ideal-gas equation to convert moles of \(\ce{CO2}\) to a volume.
\[ \begin{align*} V &= \dfrac{nRT}{PR} \\[4pt] &= \dfrac{(0.00150\;mol)\left( 0.08206\; \frac{L \cdot atm}{mol \cdot K} \right) ( 273.15\;K)}{1\;atm} \\[4pt] &= 0.0336\;L \; or \; 33.6\;mL \end{align*}\]
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.)