<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 15.7: Stoichiometry Involving Gases

Difficulty Level: At Grade Created by: CK-12

## 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 non-standard 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, 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of hydrogen gas will occupy the same volume as 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of oxygen gas if their temperatures and pressures are the same. With the same logical extension, we can say that at the same temperature and pressure, 2.00 moles\begin{align*}2.00\ \mathrm{moles}\end{align*} of oxygen gas will occupy twice the volume of 1.00 mole\begin{align*}1.00\ \mathrm{mole}\end{align*} of hydrogen gas. This logic allows us an additional way to read chemical equations involving only gases.

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: 2 molecules\begin{align*}2\ \mathrm{molecules}\end{align*} of hydrogen gas react with 1 molecule\begin{align*}1\ \mathrm{molecule}\end{align*} of oxygen gas to yield 2 molecules\begin{align*}2\ \mathrm{molecules}\end{align*} of gaseous water. After we learned about moles, we could also read this equation as: 2 moles\begin{align*}2\ \mathrm{moles}\end{align*} of hydrogen gas react with 1 mole\begin{align*}1\ \mathrm{mole}\end{align*} of oxygen gas to yield 2 moles\begin{align*}2\ \mathrm{moles}\end{align*} of gaseous water. Now, we have a third way to read this all gaseous equation: 2\begin{align*}2\end{align*} volumes of hydrogen gas react with 1\begin{align*}1\end{align*} volume of oxygen gas to yield 2\begin{align*}2\end{align*} volumes of gaseous water, if all substances are at the same temperature and pressure. The reacting ratio indicated by the coefficients in this equation are true for molecules, moles, and volumes for a reaction involving only gases as long as all the gases are under the same conditions.

Example:

What volume of oxygen gas is necessary to react with 100. L\begin{align*}100.\ \mathrm{L}\end{align*} of hydrogen gas, assuming all volumes are measured at the same temperature and pressure?

Solution:

The reacting ratio indicated by the coefficients from the equation are true for gas volumes under equal conditions.

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 2 molecules\begin{align*}2\ \mathrm{molecules}\end{align*} of hydrogen will require 1\begin{align*}1\end{align*} molecule of oxygen to react completely, so 100. L\begin{align*}100.\ \mathrm{L}\end{align*} of hydrogen will required 50.0 L\begin{align*}50.0\ \mathrm{L}\end{align*} of oxygen to react completely.

2 mol H21 mol O2=100. L of H2x L of O2     so     x=50.0 L of O2\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*}

## Volume-Volume 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 22.4 liters.\begin{align*}22.4\ \mathrm{liters.}\end{align*} If you are given a volume of gas at STP as a known, you convert it to moles by dividing by 22.4 L/mol\begin{align*}22.4\ \mathrm{L/mol}\end{align*}. If you are given a volume of gas at conditions other than STP, you must use PV=nRT\begin{align*}PV = nRT\end{align*} to calculate the number of moles.

Example:

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 50.0 L\begin{align*}50.0\ \mathrm{L}\end{align*} of nitrogen gas, N2\begin{align*}\text{N}_2\end{align*}. Assume all gases are at STP.

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.

1 N22 NH3=50.0 L of N2x L of NH3     so     x=100. L\ of NH3\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:

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 30.0 L\begin{align*}30.0\ \mathrm{L}\end{align*} of hydrogen gas at STP?

Solution:

Since the hydrogen gas volume is given at STP, we can convert the volume to moles by dividing by 22.4 L/mol\begin{align*}22.4\ \mathrm{L/mol}\end{align*}. Once we have the given quantity in moles, the remainder of the problem is the same as other stoichiometry problems.

\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*}

## Mole-Volume or Volume-Mole 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

1. 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*}?
2. 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?

All images, unless otherwise stated, are created by the CK-12 Foundation and are under the Creative Commons license CC-BY-NC-SA.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: