Volume Relationships in Gaseous Reactions Revision Notes for Grade 11 NSC Matric Physical Sciences

Volume Relationships in Gaseous Reactions

Understanding volume relationships in gases

When studying chemical reactions involving gases, we can apply the principles of stoichiometry along with our knowledge of gas behaviour. The key concept that connects these areas is molar volume - the volume that one mole of any gas occupies under standard conditions.

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Molar volume is the volume occupied by one mole of any gas at Standard Temperature and Pressure (STP).

At STP, one mole of any gas always occupies 22,4 dm³. This is a fundamental relationship that allows us to convert between the number of moles of a gas and its volume.

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Critical Concept: The molar volume of 22,4 dm³ applies to ANY gas at STP, regardless of the gas type. This universal relationship is what makes volume calculations in gaseous reactions possible.

Essential formulas for volume calculations

The two key formulas you need to master are:

Number of moles: $n = \frac{m}{M}$ where n = moles, m = mass (g), M = molar mass (g/mol)
Volume at STP: $V = 22,4 \times n$ where V = volume (dm³), n = number of moles

Step-by-step method for volume problems

Follow this systematic approach when solving volume relationship problems:

Calculate moles of the given substance using mass and molar mass
Use mole ratios from the balanced chemical equation to find moles of the required gas
Calculate volume using the molar volume relationship

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Worked Example 1: Explosive Reactions

Problem: Ammonium nitrate is used as an explosive in mining. When heated, it undergoes the following reaction:

$2NH_4NO_3(s) \rightarrow 2N_2(g) + 4H_2O(g) + O_2(g)$

If 750 g of ammonium nitrate is used, what volume of oxygen gas would be produced at STP?

Solution:

Step 1: Calculate moles of ammonium nitrate

Number of moles = $\frac{mass}{molar\ mass} = \frac{750}{80} = 9,375\ mol$

Step 2: Find moles of oxygen using mole ratios

From the balanced equation, the ratio of $NH_4NO_3$ to $O_2$ is 2:1
Moles of $O_2 = 9,375 \times \frac{1\ mol\ O_2}{2\ mol\ NH_4NO_3} = 4,6875\ mol$

Step 3: Calculate volume of oxygen

Volume = $22,4 \times 4,6875 = 105\ dm^3$

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Worked Example 2: Safety Applications

Problem: Sodium azide is used in airbags. When triggered, it produces nitrogen gas according to this reaction:

$2NaN_3(s) \rightarrow 2Na(s) + 3N_2(g)$

If 55 grams of sodium azide is used, what volume of nitrogen gas would be produced?

Solution:

Step 1: Calculate moles of sodium azide

Number of moles = $\frac{55}{65} = 0,85\ mol$

Step 2: Find moles of nitrogen using mole ratios

From the balanced equation, the ratio of $NaN_3$ to $N_2$ is 2:3
Moles of $N_2 = 0,85 \times \frac{3\ mol\ N_2}{2\ mol\ NaN_3} = 1,27\ mol$

Step 3: Calculate volume of nitrogen

Volume = $22,4 \times 1,27 = 28,4\ dm^3$

Understanding mole ratios in gaseous reactions

The stoichiometric coefficients in balanced chemical equations tell us the mole ratios between reactants and products. These ratios are crucial for calculating how much gas will be produced from a given amount of reactant.

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When working with volume relationships:

Always start with a balanced chemical equation
Use the coefficients to establish mole ratios
Remember that the molar volume relationship only applies to gases at STP

Practical applications

Volume relationships in gaseous reactions have important real-world applications:

Explosives: Understanding gas production helps predict the explosive force
Safety devices: Airbags rely on rapid gas production to inflate quickly
Industrial processes: Many manufacturing processes depend on precise gas volume calculations

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Key Points to Remember:

One mole of any gas occupies 22,4 dm³ at STP - this is the fundamental relationship for all volume calculations
Follow the three-step method: calculate moles → use mole ratios → find volume
Mole ratios come from stoichiometric coefficients in the balanced chemical equation
Always check that your chemical equation is balanced before starting calculations
Volume relationships only apply to gases - solids and liquids follow different rules

Volume Relationships in Gaseous Reactions (Grade 11 NSC Matric Physical Sciences): Revision Notes