When 200 g of lime strongly heated , it undergoes thermal decomposition to form 112 g of lime and unknown mass of carbon dioxide gas as `underset(200 g)(CaCO_(3))rarr underset(112 g)(CaO)+underset(?)(CO_(2))`
What will be the mass of `CO_(2)` formed ?

Options

A. 88 g
B. 24 g
C. 64 g
D. 40 g

Answer

To find the mass of carbon dioxide (CO₂) formed when 200 g of calcium carbonate (CaCO₃) is strongly heated, we can follow these steps:

### Step 1: Write the balanced chemical equation for the decomposition of calcium carbonate.
The thermal decomposition of calcium carbonate can be represented as:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

### Step 2: Determine the molar masses of the compounds involved.
- Molar mass of calcium carbonate (CaCO₃):
- Calcium (Ca) = 40 g/mol
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 3 = 48 g/mol
- Total = 40 + 12 + 48 = 100 g/mol

- Molar mass of carbon dioxide (CO₂):
- Carbon (C) = 12 g/mol
- Oxygen (O) = 16 g/mol × 2 = 32 g/mol
- Total = 12 + 32 = 44 g/mol

### Step 3: Use stoichiometry to find the mass of CO₂ produced from 200 g of CaCO₃.
From the balanced equation, we see that:
1 mole of CaCO₃ produces 1 mole of CO₂.

Using the molar mass of CaCO₃:
- 100 g of CaCO₃ produces 44 g of CO₂.

Now, we can set up a proportion to find the mass of CO₂ produced from 200 g of CaCO₃:
\[ \text{If } 100 \text{ g of CaCO}_3 \text{ produces } 44 \text{ g of CO}_2, \]
\[ \text{Then } 200 \text{ g of CaCO}_3 \text{ produces } x \text{ g of CO}_2. \]
Using the unitary method:
\[ x = \frac{44 \text{ g}}{100 \text{ g}} \times 200 \text{ g} = 88 \text{ g} \]

### Conclusion
The mass of carbon dioxide (CO₂) formed when 200 g of calcium carbonate (CaCO₃) is strongly heated is **88 g**.

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