22.4 litres of `H_(2)S` and 22.4 litre of `SO_(2)` both at STP are mixed together. The amount of sulphur precipitated as a result of chemical calculate is

Options

A. 16 gm
B. 23gm
C. 48gm
D. 96gm

Answer

To solve the problem of how much sulfur is precipitated when 22.4 liters of H₂S and 22.4 liters of SO₂ are mixed at STP, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between hydrogen sulfide (H₂S) and sulfur dioxide (SO₂) can be represented as follows:

\[ 2 \, \text{H}_2\text{S} + \text{SO}_2 \rightarrow 3 \, \text{S} + 2 \, \text{H}_2\text{O} \]

### Step 2: Determine the number of moles of each reactant
At STP (Standard Temperature and Pressure), 22.4 liters of any gas corresponds to 1 mole. Therefore, we have:

- Moles of H₂S = 22.4 L / 22.4 L/mol = 1 mole
- Moles of SO₂ = 22.4 L / 22.4 L/mol = 1 mole

### Step 3: Identify the limiting reactant
From the balanced equation, we see that:
- 2 moles of H₂S react with 1 mole of SO₂.

Given that we have 1 mole of H₂S and 1 mole of SO₂, we can find out how much of each reactant is needed.

For 1 mole of SO₂, we need 2 moles of H₂S. Since we only have 1 mole of H₂S, H₂S is the limiting reactant.

### Step 4: Calculate the amount of sulfur produced
From the balanced equation, 2 moles of H₂S produce 3 moles of sulfur (S).

Thus, if 2 moles of H₂S yield 3 moles of sulfur, then 1 mole of H₂S will yield:

\[ \text{Moles of S} = \frac{3}{2} \times 1 = 1.5 \, \text{moles of S} \]

### Step 5: Calculate the mass of sulfur produced
The molar mass of sulfur (S) is approximately 32 g/mol. Therefore, the mass of sulfur produced is:

\[ \text{Mass of S} = \text{Moles of S} \times \text{Molar mass of S} = 1.5 \, \text{moles} \times 32 \, \text{g/mol} = 48 \, \text{g} \]

### Final Answer
The amount of sulfur precipitated as a result of the chemical reaction is **48 grams**.