What will be the volume of` CO_(2)` at NTP obtained on heating
10 grams of (90% pure) limestone?

Options

A. `22.4` litre
B. `2.016` litre
C. `2.24` litre
D. `20.16` litre

Explanation

To find the volume of CO₂ at NTP obtained from heating 10 grams of 90% pure limestone (CaCO₃), we can follow these steps:

### Step 1: Write the balanced chemical equation
When limestone (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂). The balanced equation is:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Calculate the molar mass of CaCO₃
The molar mass of CaCO₃ can be calculated as follows:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Total molar mass of CaCO₃:
\[ 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 3: Calculate the number of moles of limestone
Since the limestone is 90% pure, the effective mass of CaCO₃ in 10 grams is:
\[ \text{Effective mass} = 10 \text{ g} \times 0.90 = 9 \text{ g} \]

Now, we can calculate the number of moles of CaCO₃:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} = \frac{9 \text{ g}}{100 \text{ g/mol}} = 0.09 \text{ moles} \]

### Step 4: Determine the volume of CO₂ produced
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.09 moles of CaCO₃ will produce 0.09 moles of CO₂.

At NTP (Normal Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Thus, the volume of CO₂ produced is:
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \text{ L/mol} = 0.09 \text{ moles} \times 22.4 \text{ L/mol} = 2.016 \text{ liters} \]

### Conclusion
The volume of CO₂ obtained from heating 10 grams of 90% pure limestone at NTP is **2.016 liters**.

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