18.4g of a mixture of `CaCO_(3)` and `MgCO_(3)` on heating gives 4.0g of magnesium oxide. The volume of `CO_(2)` produced at STP in this process is

Options

A. 1.12 L
B. 4.48 L
C. 2.24 L
D. 3.36 L

Answer

To solve the problem step by step, we will follow the thermal decomposition reactions of calcium carbonate (CaCO₃) and magnesium carbonate (MgCO₃) and calculate the volume of carbon dioxide (CO₂) produced at STP.

### Step 1: Understand the decomposition reactions
When heated, both calcium carbonate and magnesium carbonate decompose as follows:
- **Calcium Carbonate:**
CaCO₃ (s) → CaO (s) + CO₂ (g)

- **Magnesium Carbonate:**
MgCO₃ (s) → MgO (s) + CO₂ (g)

### Step 2: Determine the molar masses
- Molar mass of **Magnesium Carbonate (MgCO₃)**:
Mg (24) + C (12) + O (3 × 16) = 84 g/mol

- Molar mass of **Magnesium Oxide (MgO)**:
Mg (24) + O (16) = 40 g/mol

- Molar mass of **Calcium Carbonate (CaCO₃)**:
Ca (40) + C (12) + O (3 × 16) = 100 g/mol

### Step 3: Calculate the mass of magnesium carbonate
Given that 4.0 g of magnesium oxide is produced, we can set up a proportion to find the mass of magnesium carbonate that produced this amount.

From the reaction:
- 84 g of MgCO₃ produces 40 g of MgO.

Using the ratio:
\[ \frac{84 \text{ g MgCO}_3}{40 \text{ g MgO}} = \frac{x \text{ g MgCO}_3}{4 \text{ g MgO}} \]
Cross-multiplying gives:
\[ x = \frac{84 \times 4}{40} = 8.4 \text{ g MgCO}_3 \]

### Step 4: Calculate the mass of calcium carbonate
The total mass of the mixture is 18.4 g. Therefore, the mass of calcium carbonate can be calculated as follows:
\[ \text{Mass of CaCO}_3 = \text{Total mass} - \text{Mass of MgCO}_3 = 18.4 \text{ g} - 8.4 \text{ g} = 10.0 \text{ g} \]

### Step 5: Calculate the number of moles of each carbonate
- **Moles of Calcium Carbonate (CaCO₃)**:
\[ \text{Moles of CaCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{10.0 \text{ g}}{100 \text{ g/mol}} = 0.1 \text{ moles} \]

- **Moles of Magnesium Carbonate (MgCO₃)**:
\[ \text{Moles of MgCO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8.4 \text{ g}}{84 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 6: Calculate the total moles of CO₂ produced
From the reactions, we know that:
- 1 mole of CaCO₃ produces 1 mole of CO₂.
- 1 mole of MgCO₃ produces 1 mole of CO₂.

Thus, the total moles of CO₂ produced:
\[ \text{Total moles of CO}_2 = 0.1 + 0.1 = 0.2 \text{ moles} \]

### Step 7: Calculate the volume of CO₂ at STP
At STP, 1 mole of any gas occupies 22.4 liters. Therefore, the volume of CO₂ produced is:
\[ \text{Volume of CO}_2 = \text{Total moles} \times \text{Molar volume} = 0.2 \text{ moles} \times 22.4 \text{ L/mol} = 4.48 \text{ L} \]

### Final Answer
The volume of CO₂ produced at STP is **4.48 liters**.

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