21.75 g `MnO_(2)` is heated with HCI. The volume of `CI_(2)` obtained at NTP will be (Mn= 55, O- 16, C1= 35.5, H= 1)

Options

A. 2.8L
B. 5.6 L
C. 11.2L
D. None of these

Explanation

To solve the problem of how much chlorine gas (Cl₂) is produced when 21.75 g of manganese dioxide (MnO₂) is heated with hydrochloric acid (HCl), we can follow these steps:

### Step 1: Write the balanced chemical equation
When MnO₂ reacts with HCl, the reaction can be represented as:
\[ \text{MnO}_2 + 4 \text{HCl} \rightarrow \text{MnCl}_2 + 2 \text{H}_2\text{O} + \text{Cl}_2 \]

### Step 2: Calculate the molar mass of MnO₂
The molar mass of MnO₂ can be calculated as follows:
- Molar mass of Mn = 55 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of O₂ = 16 g/mol × 2 = 32 g/mol

Thus, the molar mass of MnO₂ is:
\[ \text{Molar mass of MnO}_2 = 55 + 32 = 87 \text{ g/mol} \]

### Step 3: Calculate the number of moles of MnO₂
Using the mass of MnO₂ given (21.75 g), we can calculate the number of moles:
\[ \text{Number of moles of MnO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{21.75 \text{ g}}{87 \text{ g/mol}} \approx 0.25 \text{ mol} \]

### Step 4: Determine the moles of Cl₂ produced
From the balanced equation, we see that 1 mole of MnO₂ produces 1 mole of Cl₂. Therefore, the moles of Cl₂ produced will also be 0.25 mol.

### Step 5: Calculate the volume of Cl₂ at NTP
At Normal Temperature and Pressure (NTP), 1 mole of gas occupies 22.4 liters. Thus, the volume of Cl₂ produced can be calculated as:
\[ \text{Volume of Cl}_2 = \text{moles of Cl}_2 \times 22.4 \text{ L/mol} = 0.25 \text{ mol} \times 22.4 \text{ L/mol} = 5.6 \text{ L} \]

### Final Answer
The volume of Cl₂ obtained at NTP is **5.6 liters**.

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