If `1.4` g of calcium oxide is formed by the complete decomposition of calcium carbonate, then the amount of calcium carbonate taken and the amount of carbon dioxide formed will be respectively :

Options

A. `2.2` g and `1.1` g
B. `1.1` g and `2.5` g
C. `2.5` g and `1.1` g
D. `5.0` g and `1.1` g

Answer

To solve the problem, we need to determine the amount of calcium carbonate (CaCO₃) that decomposes to form 1.4 g of calcium oxide (CaO) and the amount of carbon dioxide (CO₂) produced in the process.

### Step-by-Step Solution:

1. **Write the Decomposition Reaction**:
The decomposition of calcium carbonate can be represented by the following chemical equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

2. **Determine Molar Masses**:
We need the molar masses of the compounds involved:
- Molar mass of CaCO₃ = 40 (Ca) + 12 (C) + 3 × 16 (O) = 100 g/mol
- Molar mass of CaO = 40 (Ca) + 16 (O) = 56 g/mol
- Molar mass of CO₂ = 12 (C) + 2 × 16 (O) = 44 g/mol

3. **Use the Law of Conservation of Mass**:
According to the law of conservation of mass, the mass of reactants equals the mass of products. If we denote the mass of calcium carbonate taken as \( x \) and the mass of carbon dioxide produced as \( y \), we can write:
\[ x = \text{mass of CaO} + \text{mass of CO}_2 \]
Given that the mass of CaO formed is 1.4 g, we can express this as:
\[ x = 1.4 + y \]

4. **Relate the Masses Using Stoichiometry**:
From the balanced equation, we know that 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Therefore, we can set up the following relationships based on the molar masses:
- From the mass of CaO produced (1.4 g), we can find the moles of CaO:
\[ \text{Moles of CaO} = \frac{1.4 \text{ g}}{56 \text{ g/mol}} = 0.025 \text{ mol} \]
- Since the stoichiometry is 1:1, the moles of CaCO₃ decomposed is also 0.025 mol, which means:
\[ \text{Mass of CaCO}_3 = 0.025 \text{ mol} \times 100 \text{ g/mol} = 2.5 \text{ g} \]

5. **Calculate the Mass of CO₂ Produced**:
Using the moles of CaCO₃, we can find the moles of CO₂ produced:
\[ \text{Moles of CO}_2 = 0.025 \text{ mol} \]
Now, calculate the mass of CO₂:
\[ \text{Mass of CO}_2 = 0.025 \text{ mol} \times 44 \text{ g/mol} = 1.1 \text{ g} \]

6. **Final Results**:
The amount of calcium carbonate taken is **2.5 g** and the amount of carbon dioxide formed is **1.1 g**.

### Summary of Results:
- Amount of calcium carbonate taken: **2.5 g**
- Amount of carbon dioxide formed: **1.1 g**