'X' grams of calcium carbonate was completely burnt in air. The weight of the solid residue formed is 28 g. What is the value of 'X' (in grams)

Options

A. 44
B. 200
C. 150
D. 50

Explanation

To solve the problem, we need to determine the amount of calcium carbonate (CaCO₃) that was initially present based on the solid residue formed after combustion. The residue is calcium oxide (CaO), which weighs 28 grams.

### Step-by-step Solution:

1. **Write the Reaction**:
When calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂):
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

2. **Identify Molar Masses**:
- 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

3. **Determine the Moles of CaO**:
We know that the mass of CaO produced is 28 g. To find the number of moles of CaO produced, we use the formula:
\[ \text{Moles of CaO} = \frac{\text{mass}}{\text{molar mass}} = \frac{28 \text{ g}}{56 \text{ g/mol}} = 0.5 \text{ moles} \]

4. **Relate Moles of CaO to Moles of CaCO₃**:
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CaO. Therefore, the moles of CaCO₃ that decomposed to produce 0.5 moles of CaO is also 0.5 moles.

5. **Calculate the Mass of CaCO₃**:
Now, we can calculate the mass of CaCO₃ that corresponds to 0.5 moles:
\[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 0.5 \text{ moles} \times 100 \text{ g/mol} = 50 \text{ g} \]

Thus, the value of 'X' is **50 grams**.

### Summary:
- The initial mass of calcium carbonate (X) that was burnt is **50 grams**.