80 mL of pure phosphine is decomposed to produce vapours of phosphorus and `H_2`. The change in volume during reaction is :

Options

A. 60
B. 40
C. 20
D. 10

Explanation

To solve the problem of the change in volume during the decomposition of 80 mL of phosphine (PH₃) into phosphorus and hydrogen, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Decomposition Reaction:**
The decomposition of phosphine can be represented by the following balanced chemical equation:
\[ 4 \text{PH}_3 \rightarrow \text{P} + 6 \text{H}_2 \]
This equation shows that 4 moles of phosphine decompose to produce 1 mole of phosphorus and 6 moles of hydrogen.

2. **Determine the Volume Ratios:**
According to the balanced equation, for every 4 volumes of phosphine, we get:
- 1 volume of phosphorus (P)
- 6 volumes of hydrogen (H₂)

3. **Calculate the Volume of Hydrogen Produced:**
Given that we have 80 mL of phosphine, we can calculate the volume of hydrogen produced using the ratios from the balanced equation:
\[ \text{Volume of H}_2 = \left(\frac{6 \text{ mL H}_2}{4 \text{ mL PH}_3}\right) \times 80 \text{ mL PH}_3 \]
\[ \text{Volume of H}_2 = \frac{6}{4} \times 80 = 120 \text{ mL} \]

4. **Calculate the Change in Volume:**
The change in volume can be calculated by subtracting the initial volume of phosphine from the total volume of gases produced (which includes the volume of hydrogen and neglects the volume of solid phosphorus):
\[ \text{Change in Volume} = \text{Volume of H}_2 + \text{Volume of P} - \text{Volume of PH}_3 \]
Since the volume of solid phosphorus is negligible, we can simplify this to:
\[ \text{Change in Volume} = 120 \text{ mL} - 80 \text{ mL} = 40 \text{ mL} \]

5. **Final Answer:**
The change in volume during the reaction is:
\[ \text{Change in Volume} = 40 \text{ mL} \]