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 determining the amount of sulfur precipitated when 22.4 liters of H₂S and 22.4 liters of SO₂ are mixed at STP, we can 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:

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

### Step 2: Determine the moles of reactants
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Therefore, we can calculate the moles of H₂S and SO₂:

- Moles of H₂S = \( \frac{22.4 \, \text{liters}}{22.4 \, \text{liters/mole}} = 1 \, \text{mole} \)
- Moles of SO₂ = \( \frac{22.4 \, \text{liters}}{22.4 \, \text{liters/mole}} = 1 \, \text{mole} \)

### Step 3: Identify the limiting reactant
From the balanced equation, we see that 2 moles of H₂S react with 1 mole of SO₂. Therefore, 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
According to the balanced equation, 2 moles of H₂S produce 3 moles of sulfur (S). Therefore, 1 mole of H₂S will produce:

\[ \text{Moles of S} = \frac{3}{2} \times \text{moles of H}_2\text{S} = \frac{3}{2} \times 1 = 1.5 \, \text{moles of S} \]

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

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

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

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