# Gas Calculations Worksheets, Questions, and Revision

## Gas Calculations Worksheets, Questions, and Revision

**Gas Calculations**

One of the most interesting properties of **gasses** is the volume of space that they occupy. All gasses, no matter the substances, will occupy **the same volume of space per mole ** at a **given temperature and pressure**. At **20\degree \text{C}** and **1\text{ atm (atmospheres)} of pressure**, one mole of gas will occupy a volume **24\text{ dm}^3**. This will be true **no matter what substance** makes up the gas. This means that we can **always** calculate the moles of a gas (and therefore its mass) in a given volume provided we know it’s **temperature and pressure**.

**Calculating Moles and Volumes **

Calculation of the moles or volume of a gas is relatively simple. If we know one, we are able to find the other thought the following relationship:

**\text{Volume of Gas at }20\degree \text{C}\text{ and }1\text{ atm} = 24 \times \text{Moles of Gas Present}**

If we know the volume occupied by a gas at** 20\degree \text{C} and 1\text{ atm}** of pressure we can calculate the number of moles present by rearranging this formula.

It is important to remember that this relationship will only hold for gasses at **20\degree\text{C} and 1\text{ atm}**. At other temperatures or pressures, the volume occupied by one mole of gas will be **different** and so this different volume will need to be used in the above equation.

**Calculating Gas Volumes from Equations**

One result of this property of gasses is that, if we have a **balanced equation** where one or more of the chemicals involved is a gas, we can work out the **volume** of space occupied by this gas at **20\degree\text{C and }1\text{ atm}**. This because the balanced equation will tell us the **ratios of moles of each chemical are present in the reaction**.

Take the example of the reaction between hydrochloric acid \left(\text{HCl}\right) and sodium hydrogen carbonate \left(\text{NaHCO}_3\right):

\text{HCl}_{\text{ (aq)}}+\text{NaHCO}_{3\text{ (s)}} \rarr \text{NaCl}_{\text{ (s)}}+\text{H}_2\text{O}_{\text{ (l)}}+\text{CO}_{2\text{ (g)}}

This equation tells us that the ratio of moles of \left(\text{HCl}\right) & \left(\text{NaHCO}_3\right) to \text{CO}_2 is **1:1**. As such, if we know the mass of either \left(\text{HCl}\right)or \left(\text{NaHCO}_3\right), we can calculate the volume of \text{CO}_2 evolved.

**Example 1: Calculating Volume**

Calculate the volume occupied by \textcolor{#00bfa8}{2.3\text{ moles}} of nitrogen gas at 20\degree\text{C and }1\text{ atm}:

**[1 mark]**

Volume of Nitrogen at 20\degree \text{C} and 1\text{ atm}:

= 24 \times \text{Moles of Nitrogen Present}\\ = 24\times \textcolor{#00bfa8}{2.3}\\ =\textcolor{#008d65}{55.2\text{ dm}^3}

**Example 2: Calculating Moles**

Calculate the moles of oxygen present in \textcolor{#00bfa8}{179\text{ dm}^3} of gas at 20\degree\text{C and }1\text{ atm}:

**[1 mark]**

Volume of Oxygen at 20\degree \text{C} and 1\text{ atm}:

= 24 \times \text{Moles of Oxygen Present}

\text{Moles of Oxygen} =\frac{\text{Volume of Oxygen}}{24}\\ \, \\= \frac{\textcolor{#00bfa8}{179}}{24}\\ \, \\ =\textcolor{#008d65}{7.5\text{ mol}}

**Example 3: Calculating Volumes from Equations**

Calculate the volume of \text{CO}_2 released when \textcolor{#00bfa8}{8.35\text{ g}} of \text{NaHCO}_3_{ }(\text{M}_r =\textcolor{#f21cc2}{84}) reacts with \text{HCl} at 20\degree\text{C and }1\text{ atm}:

**[3 marks]**

Step 1: calculate the moles of \text{NaHCO}_3 that have reacted:

\text{Moles NaHCO}_3 = \frac{\text{Mass NaHCO}_3}{\text{M}_r \text{NaHCO}_3}\\ \, \\ =\frac{\textcolor{#00bfa8}{8.35}}{\textcolor{#f21cc2}{84}}\\ \, \\=\textcolor{#008d65}{0.099\text{ mol}}

Step 2: calculate the moles of \text{CO}_2 that are evolved from the balanced equation of the reaction:

\text{HCl}_{\text{ (aq)}}+\text{NaHCO}_{3\text{ (s)}} \rarr \text{NaCl}_{\text{ (s)}}+\text{H}_2\text{O}_{\text{ (l)}}+\text{CO}_{2\text{ (g)}}

\text{Ratio of Moles}=1:1

\text{Moles of CO}_2=\text{Moles of NaHCO}_3

Step 3: calculate the volume occupied by \text{CO}_2 at 1\degree\text{C and }1\text{ atm}:

Volume of \text{CO}_2 at 20\degree \text{C} and 1\text{ atm}:

= 24 \times \text{Moles of CO}_2\\ \, \\= 24\times 0.099\\ \, \\=\textcolor{#008d65}{2.4\text{ dm}^3}

## Gas Calculations Worksheets, Questions, and Revision Example Questions

**Question 1: **Calculate the volume of gas occupied by 0.070\text{ mol} of HCl Gas at 20\degree\text{C and }1\text{ atm}.

**[1 mark]**

\begin{aligned}\text{Volume of HCl at }20\degree \text{C}\text{ and }1\text{ atm} &= 24 \times \text{Moles of HCl Present}\\ &=24 \times 0.070 \\ &=\underline{1.68\text{ dm}^3}\end{aligned}

**Question 2: **Calculate the moles of air present in a room of volume 70\text{ million cm}^3 at 20\degree\text{C and }1\text{ atm}.

**[2 marks]**

Firstly, convert the volume of the room into \text{ dm}^3:

\begin{aligned}\text{Volume in dm}^3&=\frac{\text{Volume in cm}^3}{1000}\\ &=\frac{70000000}{1000}\\ &=\underline{70000\text{ dm}^3}\end{aligned}

Second, calculate the moles of air present:

\begin{aligned}\text{Moles of Air}&=\frac{\text{Volume of Room}}{24}\\ &=\frac{70000}{24}\\ &=\underline{2917\text{ mol}}\end{aligned}

**Question 3: **Calculate the volume of \text{CO}_2 released when 4.67\text{ g} of \text{NaHCO}_3(\text{M}_r =84) reacts with \text{HCl} at 20\degree\text{C and } 1\text{ atm}.

**[3 marks]**

Calculation should be broken down into steps. One mark per correct step.

Balanced Equation:

\text{HCl}_{\text{ (aq)}}+\text{NaHCO}_{3\text{ (s)}} \rarr \text{NaCl}_{\text{ (s)}}+\text{H}_2\text{O}_{\text{ (l)}}+\text{CO}_{2\text{ (g)}}

Step 1: Calculate moles of \text{NaHCO}_3.

\begin{aligned}\text{Moles NaHCO}_3 &= \frac{\text{Mass NaHCO}_3}{\text{M}_r \text{NaHCO}_3}\\ &=\frac{4.67}{84}\\ &=\underline{0.056\text{ mol}}\end{aligned}

Step 2: Calculate Moles of \text{CO}_2.

\text{Ratio of Moles}=1:1

\text{Moles of CO}_2=\text{Moles of NaHCO}_3

Step 3: Calculate the volume of \text{CO}_2\text{ at } 20\degree\text{C and }1\text{ atm}:

\begin{aligned}\text{Volume of CO}_2\text{ at }20\degree \text{C}\text{ and }1\text{ atm} &= 24 \times \text{Moles of CO}_2\\ & 24\times 0.056\\ &=\underline{1.34\text{ dm}^3}\end{aligned}