Mass of `CO_(2)` Produced on heating 20g of 40% pure limestone :-

Options

A. 8gm
B. 8.8gm
C. 3.52gm
D. none of these

Explanation

To solve the problem of finding the mass of CO₂ produced from heating 20 grams of 40% pure limestone (CaCO₃), we can follow these steps:

### Step 1: Determine the mass of pure CaCO₃ in the limestone sample
Given that the limestone is 40% pure, we can calculate the mass of pure CaCO₃ in the 20 grams of limestone.

\[ \text{Mass of pure CaCO}_3 = \text{Total mass of limestone} \times \text{Purity} \]
\[ \text{Mass of pure CaCO}_3 = 20 \, \text{g} \times \frac{40}{100} = 8 \, \text{g} \]

### Step 2: Write the balanced chemical equation for the decomposition of CaCO₃
When limestone (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂):

\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 3: Calculate the molar mass of CaCO₃ and CO₂
- Molar mass of CaCO₃:
- Ca: 40 g/mol
- C: 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + 48 = 100 \, \text{g/mol} \]

- Molar mass of CO₂:
- C: 12 g/mol
- O: 16 g/mol × 2 = 32 g/mol
\[ \text{Molar mass of CO}_2 = 12 + 32 = 44 \, \text{g/mol} \]

### Step 4: Use stoichiometry to find the mass of CO₂ produced
From the balanced equation, 100 grams of CaCO₃ produces 44 grams of CO₂. We can set up a proportion to find out how much CO₂ is produced from 8 grams of CaCO₃.

\[ \text{If } 100 \, \text{g CaCO}_3 \text{ produces } 44 \, \text{g CO}_2 \]
\[ \text{Then } 8 \, \text{g CaCO}_3 \text{ produces } x \, \text{g CO}_2 \]
Using the unitary method:
\[ x = \frac{44 \, \text{g CO}_2}{100 \, \text{g CaCO}_3} \times 8 \, \text{g CaCO}_3 \]
\[ x = \frac{44 \times 8}{100} = 3.52 \, \text{g CO}_2 \]

### Final Answer
The mass of CO₂ produced when heating 20 grams of 40% pure limestone is **3.52 grams**.

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