# 9.4: Gas Phase Stoichiometry

- Page ID
- 284094

## Introduction

Now that we have covered the ideal gas law we can revisit the stoichiometric equations of Chapter 4 and account for gas phase reactants and products in a more meaningful manner. While going over this material you may wish to review the following sections.

- 3.2: Types of Chemical Reactions (combustion and many types of single displacement involve gasses)
- 3.5.4: Acid Reactions Evolving Gases (Strong acid reactions with salts of oxyanions [acid anhydrides])
- 4.1: Stoichiometry
- 4.2: Limiting & Excess Reagents

In the previous sections we would report gasses in terms of their moles, or even their mass. Lets take a second look at Exercise 4.2.1, where we calculated the mass of carbon dioxide formed upon the combustion of 25.00g glucose with 40.0 g of oxygen. Sure we could solve the problem, but at standard conditions oxygen or carbon dioxide are gasses not solids, and so you would not be able to directly measure their mass, and a more realistic problem would be to define them in terms of measurable quantities like pressure, temperature and volume.

## Overview of Stoichiometry:

In stoichiometric calculations you typically have three steps.

- Convert measured quantities to moles (assuming you don't start with moles, but something measurable)
- Use balanced equations to relate moles of starting material to moles of desired substance sought.
- Convert moles of desired substance to measurable quantities (assuming you don't want moles, but something measurable)

mass | solute in solution | |

n=\(\frac{m}{fw}\) | n=MV | n=\(\frac{PV}{RT}\) |

m=mass |
M=Molarity=\(\frac{mol_{solute}}{liter_{solution}}\) |
P=Pressure |

As in Chapter 4, it is a good idea to write out and balance the equation, write under it what is given, and then solve the problem. The following problems involves calculations involving all three phases.

Example \(\PageIndex{1}\)

What volume of gas would be produced at 760.0 torr and 298 K if 1.00 g of magnesium metal is dropped into 50.0 ml of 1.00 M Hydrochloric Acid?

**Solution**

(See Video \(\PageIndex{1}\) for solution)

**Step 1:** Write out and balance equation and write underneath it what is given:

\[\underset{m=1.00g \\fw= 24.3\frac{g}{mol}}{Mg(s)} \; + \; \underset{0.0500L \\ 1.00 M}{2HCl(aq)} \; \rightarrow \; Mg(Cl)_2(aq)+\underset{P=760.0 atm \\ T=298K \\n=? \\ V=??}{H_2(g)}\]

NOTE: We want \(V_{H_2}\), but have two unknowns (V and n), so we need to calculate n

**Step 2:** Calculate theoretical yield of n (based on complete consumption of limiting reagent)

\[1.00g \; Mg\left ( \frac{mol \; Mg}{24.3g} \right )\left ( \frac{1}{1mol \; Mg} \right )=0.0411 \\ 0.0500L\left ( \frac{1.00mol \;HCl}{L} \right )\left ( \frac{1}{2mol \;HCl} \right )=.025

\\HCl =limiting \; reagent \\n_{H_2} =0.025\left ( 1mol \; H_2(g) \right )=0.025 mol\;H_2(g)\]

**Step 3: **Now that we have \(n_{H_2}\), use ideal gas law to solve for V at the conditions of the problem.

\[V=\frac{nRT}{P}=\frac{0.02500mol\left (0.08206\left ( \frac{l \cdot atm}{mol \cdot K} \right ) \right )298K}{760 \; torr\left ( \frac{1atm}{760 \;torr} \right )} = .6113L=611mL\]

## Test Yourself

**Homework:** Section 9.4

**Graded Assignment: **Section 9.4

Exercise \(\PageIndex{1}\)

What volume of oxygen is formed at 750.0torr and 25.0^{o}C if 1.30 g potassium chlorate decomposes into potassium chloride and oxygen?

**Answer**-
First write out balanced equation and calculate moles oxygen formed

\[\underset{m=1.30g \\fw= 122.55 \frac{g}{mol}}{2KClO_3(s)} \; \rightarrow \; 2KCl(s) \; + \; \underset{P=750 torr \\ T=298K \\n=? \\ V=??}{3O_2(g)}\]

First calculate moles oxygen produced by the complete decomposition of the potassium chlorate

\[1.30gKClO_3\left ( \frac{mol \; KClO_3}{122.55g} \right )\left ( \frac{3mol\;O_2}{2mol \; KClO_3} \right )=0.0159mol \; O_2\]

Substitute into ideal gas law and solve

\[V=\frac{nRT}{P}=\frac{0.0159mol\left ( 0.08206 \frac{l \cdot atm}{mol \cdot K} \right )298K}{750torr \left ( \frac{1atm}{760torr} \right )} =394mL\]

Exercise \(\PageIndex{2}\)

Solve the following problems

- 5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A single displacement reaction occurs in a sealed container at 25.0
^{o}C. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water? - Nitrogen dioxide is formed in a closed rigid container at 90
^{o}C and 1.00 atm when 1.50 g NO and 2.00 mole of O_{2}are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature?

**Answer a**-
First you need to calculate the theoretical yield based on the complete consumption of the limiting reagent.

\[5g \; Mg\left ( \frac{mol \; Mg}{24.31g} \right )\frac{1}{1mol Mg}=0.206 \\ 0.0500L\left ( \frac{0.800\;mol \;HCl}{L} \right )\frac{1}{2 \; mol \;HCl}=0.0200\] so HCl is the limiting reagent and the theoretical yield is

\[0.0500L\left ( \frac{0.800\;mol \;HCl}{L} \right )\frac{1\;mol\; H_2}{2 \; mol \;HCl}=0.0200\;mol\;H_2\]

Substituting into the ideal has equation gives the pressure\[P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\]

**Answer b**-
This combines stoichiometry with the two state approach. First write out and balance the equation, indicating initial reactants

\[\underset{m=1.50g \\fw= 30.01 \frac{g}{mol}}{2NO(g)} \; + \; \underset{2.00 \; mol}{O_2(g)} \; \rightarrow \; \underset{P=1.00 atm \\ T=373.15K \\n=? \\ V=??}{2NO_2(g)}\]

Now set up your two state table:

Initial State (1) Final State (2) Temp 90+273.15=373.15K T _{2}=?Pressure 1.00 atm 3.90 atm moles (n) n _{1}=n _{2}=Vol V _{1}=V_{2}V _{1}=V_{2}First calculate n

_{1}, the initial number of moles of all species (before reaction):\[n_1=n_{NO}+n_{O_2}+n_{NO_2}=1.50g\left ( \frac{mol\;NO}{30.01g} \right )+2.00mol \; O_2+0.00mol \; NO_2\\=0.04998mol \;NO+2.00mol \; O_2=2.05mol_{total}\]

Second calculate n

_{2}, the total number of moles all species after reactionNO is the limiting reagent because

\[\frac{0.05}{2}< \frac{2.00}{1}\]

the amount of NO

_{2}produced is:\[0.05mol\;NO \left ( \frac{2mol \; NO_2}{2mol \; NO} \right )=0.05mol \; NO_2\]

the amount of excess oxygen is

\[O_2(excess) \; = \; O_2(initial) -O_2(consumed) \\ =2.00mol-0.05mol \; NO \left ( \frac{mol \; O_2}{2 mol \; NO} \right )=2.00-0.025=1.975\]

total moles after reaction

\[n_2=n_{NO}+n_{O_2}+n_{NO_2}=0+1.975+0.05=2.025\]

\[T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\]

## Contributors and Attributions

- Bob Belford (UALR) and November Palmer (UALR)