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)?

Answer

To solve the problem, we need to determine the mass of calcium carbonate (CaCO₃) that was completely burnt to yield a solid residue of calcium oxide (CaO) weighing 28 grams.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction of calcium carbonate when burnt in air can be represented as:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]
Here, calcium carbonate decomposes to form calcium oxide and carbon dioxide.

2. **Determine Molar Masses**:
- Molar mass of calcium carbonate (CaCO₃):
\[ \text{Ca} = 40.08 \, \text{g/mol} \\ \text{C} = 12.01 \, \text{g/mol} \\ \text{O} = 16.00 \, \text{g/mol} \times 3 = 48.00 \, \text{g/mol} \\ \text{Total} = 40.08 + 12.01 + 48.00 = 100.09 \, \text{g/mol} \]
- Molar mass of calcium oxide (CaO):
\[ \text{Ca} = 40.08 \, \text{g/mol} \\ \text{O} = 16.00 \, \text{g/mol} \\ \text{Total} = 40.08 + 16.00 = 56.08 \, \text{g/mol} \]

3. **Set Up the Stoichiometric Relationship**:
From the balanced equation, we know that 1 mole of CaCO₃ produces 1 mole of CaO. Therefore, the mass of CaO produced can be used to find the mass of CaCO₃ that was burnt.

4. **Calculate Moles of CaO Produced**:
Using the mass of CaO produced:
\[ \text{Moles of CaO} = \frac{\text{mass}}{\text{molar mass}} = \frac{28 \, \text{g}}{56.08 \, \text{g/mol}} \approx 0.498 \, \text{mol} \]

5. **Calculate Moles of CaCO₃ Required**:
Since the stoichiometry of the reaction shows that 1 mole of CaCO₃ produces 1 mole of CaO, the moles of CaCO₃ required will also be approximately 0.498 mol.

6. **Calculate Mass of CaCO₃**:
Now, we can calculate the mass of CaCO₃ that corresponds to the moles calculated:
\[ \text{Mass of CaCO}_3 = \text{moles} \times \text{molar mass} = 0.498 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 49.8 \, \text{g} \]

Thus, the value of \( x \) (the mass of calcium carbonate) is approximately **49.8 grams**.