Find the volume of sulphur dioxide (at STP) that would be liberated by roasting 30 g of iron pyrites according to the equation `4FeS_(2) + 11O_(2) to 2Fe_(2)O_(3) + 8SO_(2)`.
(Atomic mass : `S = 32, Fe = 56, O = 16,` molar volume of gas is `22.4` litres at STP.)

Explanation

To find the volume of sulfur dioxide (SO₂) that would be liberated by roasting 30 g of iron pyrites (FeS₂) according to the reaction:

\[ 4 \text{FeS}_2 + 11 \text{O}_2 \rightarrow 2 \text{Fe}_2\text{O}_3 + 8 \text{SO}_2 \]

we will follow these steps:

### Step 1: Calculate the molar mass of iron pyrites (FeS₂)

The molar mass of FeS₂ can be calculated as follows:
- Atomic mass of Fe = 56 g/mol
- Atomic mass of S = 32 g/mol

Since there are 2 sulfur atoms in FeS₂:
\[ \text{Molar mass of FeS}_2 = 56 + (2 \times 32) = 56 + 64 = 120 \text{ g/mol} \]

### Step 2: Calculate the number of moles of FeS₂ in 30 g

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

We can find the number of moles of FeS₂:
\[ \text{Number of moles of FeS}_2 = \frac{30 \text{ g}}{120 \text{ g/mol}} = 0.25 \text{ moles} \]

### Step 3: Determine the moles of SO₂ produced

From the balanced chemical equation, we see that:
- 4 moles of FeS₂ produce 8 moles of SO₂.

Thus, 1 mole of FeS₂ produces 2 moles of SO₂. Therefore, we can calculate the moles of SO₂ produced from 0.25 moles of FeS₂:
\[ \text{Moles of SO}_2 = 0.25 \text{ moles FeS}_2 \times \frac{8 \text{ moles SO}_2}{4 \text{ moles FeS}_2} = 0.25 \times 2 = 0.5 \text{ moles SO}_2 \]

### Step 4: Calculate the volume of SO₂ at STP

At STP (Standard Temperature and Pressure), the molar volume of any gas is 22.4 liters. Therefore, the volume of 0.5 moles of SO₂ can be calculated as:
\[ \text{Volume of SO}_2 = \text{Number of moles} \times \text{Molar volume} \]
\[ \text{Volume of SO}_2 = 0.5 \text{ moles} \times 22.4 \text{ L/mol} = 11.2 \text{ liters} \]

### Final Answer
The volume of sulfur dioxide liberated by roasting 30 g of iron pyrites is **11.2 liters**.

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