## Calculations For Chemical Reactions Involving Gases

### Molar gas volume

The volume of 1 mole of any gas is known as its molar gas volume. The molar gas volume is the same for all gases because equal volumes of different gases have the same number of moles at the same temperature and pressure.

One most important thing to note about molar gas volume is that it varies with temperature and pressure. For example, in cooler countries like the United Kingdom the molar gas volume is taken to be 24 dm3 but in warmer countries like Zimbabwe the molar gas volume is taken to be 28 dm3. This means that 1 mole of any gas occupies:

• 24 dm3 in UK
• 28 dm3 in Zimbabwe.

If you are a student, check with your teacher on the local value of molar gas volume in your country so that you wont use the wrong value in the exam. Textbooks used in most countries were printed in the UK, and in those textbooks you will find 24 dm3 being used as the molar gas volume.

### Calculating the number of moles of a gas in a certain volume at rtp

$$Number\ of\ moles = \frac{volume}{molar\ volume}$$

#### Question

Nitrogen monoxide is oxidised by oxygen as follows:

2NO(g) + O2(g) → 2NO2(g)

1. How many moles of oxygen molecules react with 1 mole of nitrogen monoxide molecules?
2. What volume of oxygen will react with 50 cm3 of nitrogen monoxide? and the volume of nitrogen dioxide is formed?

2NO(g) + O2(g) → 2NO2(g)

Looking at the reaction we can see that 2 moles of NO gas react with 1 mole of O2 gas to produce 2 moles of NO2 gas.

1. Using the ratio above, 0.5 moles of oxygen molecules react with 1 mole of nitrogen monoxide molecules.
2. Since we know that equal volumes of gases have the same number of moles, we can deduce that the ratio of the volumes of the gases is equal to their molar ratio. So, if 2 moles of NO gas react with 1 mole of O2 gas to produce 2 moles of NO2 gas this mean that 2 volumes of NO gas react with 1 volume of O2 gas to produce 2 volumes of NO2 gas.

Therefore, 50 cm3 of NO gas react with 25 cm3 of O2 gas to produce 50 cm3 moles of NO2 gas.

Volume of oxygen = 25 cm3

Volume of nitrogen dioxide = 50 cm3

### Calculating the volume that a gas will occupy at rtp given its mass, or number of moles

$$Volume = Number\ of\ moles \times Molar\ volume$$

#### Example – What is the volume of 0.25 moles of a oxygen gas occupy at rtp? Take the molar gas to be 24 dm3 at rtp.

Volume = Number of moles × Molar volume

= 0.25 moles × 24 dm3

= 6 dm3

### Calculating the volume of gas produced in a reaction, given the equation and the mass of one substance

When given mass of a reactant and asked to calculate the volume of a gas produced you follow the following steps:

• convert the mass of the reactant to moles
• use the molar ratio shown by the chemical equation to find the number of moles of gas produced by the corresponding number of moles of the reactant
• convert the number of moles of gas to volume.

#### Question

Nitroglycerine is an unstable compound that undergoes explosive decomposition as follows:

4C3H5(NO3)3(l) → 12CO2(g) + 10H2O(l) + 6N2(g) + O2(g)

1. According to the equation, what is molar ratio of nitroglycerine to the gas molecules produced?
2. What is the mass of 1 mole of nitroglycerine?
3. What is the total volume of gas (at rtp) obtained from 1 mole of nitroglycerine?
[Take the molar gas volume at rtp to be 24 dm3]

4C3H5(NO3)3(l) → 12CO2(g) + 10H2O(l) + 6N2(g) + O2(g)

1. According to the state symbols of the equation, three gases are produced: 12CO2(g), 6N2(g) and O2(g). So the total number of moles of gas produced = 12 + 6 + 1 = 19 moles. The moles of nitroglycerine = 4.

Molar ratio = 4:19

2. Mass of 1 mole of nitroglycerine = molar mass of nitroglycerine

= (12 × 3) + (1 × 5) + (14 × 3) + (16 × 9)

= 227 g

3. According to the molar ratio of 4:19

1 mole of nitroglycerine produces 19/4 = 4.75 moles of gas.

volume of gas = number of moles × molar gas volume

= 4.75 × 24

= 114 dm3