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

Options

A. 70mL
B. 60mL
C. 40mL
D. 30mL

Explanation

To solve the problem of determining the change in volume when 140 mL of pure phosphine (PH₃) decomposes to produce phosphorus (P) and hydrogen (H₂), we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of phosphine.
The decomposition of phosphine can be represented by the following balanced equation:

\[ 4 \text{PH}_3 \rightarrow \text{P} + 6 \text{H}_2 \]

### Step 2: Determine the volume ratios from the balanced equation.
From the balanced equation, we see that:
- 4 volumes of phosphine (PH₃) produce 1 volume of phosphorus (P) and 6 volumes of hydrogen (H₂).

### Step 3: Calculate the total volume of products formed from 140 mL of phosphine.
Using the volume ratios:
- If 4 mL of PH₃ produces 1 mL of P and 6 mL of H₂, then for 140 mL of PH₃, we can set up the following calculations:

1. Calculate the amount of hydrogen produced:
\[ \text{Volume of H}_2 = \left(\frac{6 \text{ mL H}_2}{4 \text{ mL PH}_3}\right) \times 140 \text{ mL PH}_3 = \frac{6}{4} \times 140 = 210 \text{ mL H}_2 \]

2. Calculate the amount of phosphorus produced:
\[ \text{Volume of P} = \left(\frac{1 \text{ mL P}}{4 \text{ mL PH}_3}\right) \times 140 \text{ mL PH}_3 = \frac{1}{4} \times 140 = 35 \text{ mL P} \]

### Step 4: Calculate the total volume of products.
The total volume of products (P and H₂) formed is:
\[ \text{Total Volume} = \text{Volume of P} + \text{Volume of H}_2 = 35 \text{ mL P} + 210 \text{ mL H}_2 = 245 \text{ mL} \]

### Step 5: Determine the change in volume.
The change in volume during the reaction can be calculated as:
\[ \text{Change in Volume} = \text{Total Volume of Products} - \text{Initial Volume of PH}_3 \]
\[ \text{Change in Volume} = 245 \text{ mL} - 140 \text{ mL} = 105 \text{ mL} \]

### Final Answer:
The change in volume during the reaction is **105 mL**.

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