15.7: Stoichiometry Involving Gases
Lesson Objectives
The student will:
 solve stoichiometry problems involving the conversion of gas volumes at STP to moles and vice versa
 solve stoichiometry problems involving the conversion of gas volumes at STP to gas volumes under nonstandard conditions
Introduction
The knowledge of the gas laws and the molar volume of gases at STP allow us to solve stoichiometry problems from gas volumes as well as masses.
Volumes, Moles, and Molecules
Recall that Avogadro’s law tells us that under the same conditions of temperature and pressure, equal volumes of gases will contain equal numbers of molecules. By logical extension, we can say that under the same conditions of temperature and pressure, equal numbers of molecules will occupy equal volumes. For example, \begin{align*}1.00\ \mathrm{mole}\end{align*}



\begin{align*}2 \ \text{H}_{2(g)} + \ \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(g)}\end{align*}
2 H2(g)+ O2(g)→2 H2O(g)

\begin{align*}2 \ \text{H}_{2(g)} + \ \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(g)}\end{align*}

The first way we learned to read this equation was: \begin{align*}2\ \mathrm{molecules}\end{align*}
Example:
What volume of oxygen gas is necessary to react with \begin{align*}100.\ \mathrm{L}\end{align*}
Solution:
The reacting ratio indicated by the coefficients from the equation are true for gas volumes under equal conditions.



\begin{align*}2 \ \text{H}_{2(g)} + \ \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(g)}\end{align*}
2 H2(g)+ O2(g)→2 H2O(g)

\begin{align*}2 \ \text{H}_{2(g)} + \ \text{O}_{2(g)} \rightarrow 2 \ \text{H}_2\text{O}_{(g)}\end{align*}

The coefficients of the balanced equation indicate that \begin{align*}2\ \mathrm{molecules}\end{align*}
\begin{align*}\frac {2\ \text{mol H}_2} {1\ \text{mol O}_2} = \frac {100.\ \text{L of H}_2} {x \ \text{L of O}_2} \ \ \ \ \ \text{so} \ \ \ \ \ x = 50.0\ \text{L of}\ O_2\end{align*}
VolumeVolume Calculations at STP
When you first learned to solve stoichiometry problems, one of the steps was to convert known quantities to moles. In those early problems, the known quantity was given in mass and you converted mass to moles by dividing grams by molar mass. From now on, you may be given known quantities of a gas as a volume at either STP or some other temperature and pressure conditions. You already know how to convert a volume of gas at STP to moles. At STP, one mole of any gas occupies \begin{align*}22.4\ \mathrm{liters.}\end{align*}
Example:



\begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}
N2(g)+3 H2(g)→2 NH3(g)

\begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}

According to this equation, how many liters of ammonia can be formed from \begin{align*}50.0\ \mathrm{L}\end{align*}
Solution:
Since both the given substance and the requested substance are gases under the same conditions of temperature and pressure, the reacting ratio indicated by the coefficients in the equation are true for volumes.
\begin{align*}\frac {1\ \text{N}_2} {2\ \text{NH}_3} = \frac {50.0\ \text{L of N}_2} {x \ \text{L of NH}_3} \ \ \ \ \ \text{so} \ \ \ \ \ x = 100.\ \text{L\ of NH}_3\end{align*}
Example:



\begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}
N2(g)+3 H2(g)→2 NH3(g)

\begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}

According to this equation, how many grams of ammonia can be formed from \begin{align*}30.0\ \mathrm{L}\end{align*}
Solution:
Since the hydrogen gas volume is given at STP, we can convert the volume to moles by dividing by \begin{align*}22.4\ \mathrm{L/mol}\end{align*}
\begin{align*}(30.0\ \text{L}) \cdot \left (\frac {1.00\ \text{mol}} {22.4\ \text{L}}\right ) = 1.34\ \text{mol H}_2\end{align*}
\begin{align*}\frac {3\ \text{H}_2} {2\ \text{NH}_3} = \frac {1.34\ \text{mol H}_2} {x\ \text{mol NH}_3} \ \ \ \ \ \text{so} \ \ \ \ \ x = 0.893\ \text{mol NH}_3\end{align*}
\begin{align*}\text{grams NH}_3 = \ \text{(moles)} \cdot \text{(molar mass)} = (0.893\ \text{mol)} \cdot (17.0\ \text{g/mol}) = 15.2\ \text{grams}\end{align*}
Example:


 \begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}

According to this equation, how many liters of ammonia gas at STP can be formed from \begin{align*}25.0\ \mathrm{g}\end{align*} of hydrogen gas?
Solution:
The solution steps are to convert the grams of \begin{align*}\text{H}_2\end{align*} to moles, use the ratio from the equation to solve for moles of ammonia gas, and then convert the moles of ammonia to liters of gas at STP. The moles of ammonia gas is converted to liters at STP by multiplying by \begin{align*}22.4 \ \mathrm{L/mol}\end{align*}.
\begin{align*}(25.0\ \text{g}) \cdot \left (\frac {1.00\ \text{mol}} {2.02\ \text{g}}\right ) = 12.4\ \text{mol}\ \text{H}_2 \end{align*}
\begin{align*}\frac {3\ \text{H}_2} {2\ \text{NH}_3} = \frac {12.4\ \text{mol H}_2} {x \ \text{mol NH}_3} \ \ \ \ \ \text{so} \ \ \ \ \ x = 8.27\ \text{mol}\ \text{NH}_3\end{align*}
\begin{align*}\text{liters}\ \text{NH}_3 = \ \text{(moles)} \cdot \text{(molar mass)} = (8.27\ \text{mol}) \cdot (22.4\ \text{L/mol}) = 185\ \text{liters}\end{align*}
MoleVolume or VolumeMole Calculations Not at STP
When volumes involved in chemical reactions are not given at standard conditions, the conversion between moles of gas and volume of gas can be determined by using the universal gas law, \begin{align*}PV = nRT\end{align*}.
Example:
\begin{align*}1,000\end{align*}. grams of calcium carbonate are heated and react according to the following equation:


 \begin{align*}\text{CaCO}_{3(s)} \rightarrow \text{CaO}_{(s)} + \text{CO}_{2(g)}\end{align*}

If the carbon dioxide is collected at \begin{align*}500.\ \mathrm{K}\end{align*} and \begin{align*}2.00\ \mathrm{atm}\end{align*}, what volume will it occupy under these conditions?
Solution:
\begin{align*}(1000.\ \text{g}) \cdot \left (\frac {1.00\ \text{mol}} {100.\ \text{g}}\right ) = 10.0\ \text{mol}\ \text{CaCO}_3\end{align*}
Since the reacting ratio between calcium carbonate and carbon dioxide in the equation is \begin{align*}1\end{align*} to \begin{align*}1\end{align*}, the \begin{align*}10.0\ \mathrm{mol}\end{align*} of calcium carbonate will form \begin{align*}10.0\ \mathrm{mol}\end{align*} of carbon dioxide. We use \begin{align*}PV = nRT\end{align*} to find the volume that \begin{align*}10.0\ \mathrm{mol}\end{align*} of gas will occupy under these conditions.
\begin{align*}V = \frac {nRT} {P} = \frac {(10.0\ \text{mol}) \cdot (0.08206\ \text{L} \cdot \text{atm} \cdot \text{mol}^{1} \cdot \text{K}^{1}) \cdot (500.\ \text{K})} {(2.00\ \text{atm})} = 205\ \text{liters}\end{align*}
Lesson Summary
 In a balanced gaseous equation, the coefficients apply to molecules, moles, and volumes of gas if the gases are under the same conditions of temperature and pressure.
Review Questions
 How many liters of hydrogen gas are required to react with \begin{align*}25.0\ \mathrm{L}\end{align*} of nitrogen gas according to the following equation: \begin{align*}\text{N}_{2(g)} + 3 \ \text{H}_{2(g)} \rightarrow 2 \ \text{NH}_{3(g)}\end{align*}?
 How many grams of ammonia will be formed from \begin{align*}25.0\ \mathrm{L}\end{align*} of nitrogen gas measured at STP according to the equation in problem #2?
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