Chlorine gas is prepared by reaction of `H_(2)SO_(4)` with `MnO_(2)` and `NaCl`. What volume of `Cl_(2)`will be produced at STP if 50 g of `NaCl` is taken in the reaction?

Options

A. 19.14 L
B. 22.4 L
C. 11.2 L
D. 9.57 L

Answer

To solve the problem of how much chlorine gas (Cl₂) will be produced from the reaction of sodium chloride (NaCl) with sulfuric acid (H₂SO₄) and manganese dioxide (MnO₂), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
\[ 2 \text{NaCl} + 3 \text{H}_2\text{SO}_4 + \text{MnO}_2 \rightarrow 2 \text{NaHSO}_4 + \text{Cl}_2 + \text{MnO}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Calculate the number of moles of NaCl
First, we need to find the molar mass of NaCl:
- Sodium (Na) = 23 g/mol
- Chlorine (Cl) = 35.5 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Now, we can calculate the number of moles of NaCl in 50 g:
\[ \text{Number of moles of NaCl} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \text{ g}}{58.5 \text{ g/mol}} \approx 0.8547 \text{ mol} \]

### Step 3: Use stoichiometry to find moles of Cl₂ produced
From the balanced equation, we see that 2 moles of NaCl produce 1 mole of Cl₂. Therefore, the number of moles of Cl₂ produced can be calculated as follows:
\[ \text{Moles of Cl}_2 = \frac{0.8547 \text{ mol NaCl}}{2} \approx 0.4274 \text{ mol Cl}_2 \]

### Step 4: Calculate the volume of Cl₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of Cl₂ produced can be calculated as:
\[ \text{Volume of Cl}_2 = \text{moles of Cl}_2 \times 22.4 \text{ L/mol} = 0.4274 \text{ mol} \times 22.4 \text{ L/mol} \approx 9.57 \text{ L} \]

### Final Answer
The volume of Cl₂ produced at STP from 50 g of NaCl is approximately **9.57 liters**.

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