18.4 gms of a mixture of calcium carbonate and magnesium carbonate on heating gives 4.0 gms of magnesium oxide. The volume of `CO_(2)` produced at STP in this process is

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the reactions for the decomposition of calcium carbonate and magnesium carbonate.
The thermal decomposition reactions are:
1. \( \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \)
2. \( \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \)

### Step 2: Determine the moles of magnesium oxide produced.
Given that 4.0 g of magnesium oxide (MgO) is produced, we can calculate the number of moles of MgO using its molar mass.
- Molar mass of MgO = Atomic mass of Mg (24 g/mol) + Atomic mass of O (16 g/mol) = 40 g/mol.

Using the formula for moles:
\[ \text{Moles of MgO} = \frac{\text{mass}}{\text{molar mass}} = \frac{4.0 \, \text{g}}{40 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 3: Relate the moles of MgO to the moles of MgCO3.
From the stoichiometry of the reaction, 1 mole of MgCO3 produces 1 mole of MgO. Therefore, the moles of MgCO3 that decomposed is also 0.1 moles.

### Step 4: Calculate the mass of MgCO3.
The molar mass of MgCO3 is calculated as follows:
- Molar mass of MgCO3 = Atomic mass of Mg (24 g/mol) + Atomic mass of C (12 g/mol) + 3 × Atomic mass of O (16 g/mol) = 24 + 12 + 48 = 84 g/mol.

Now, we can find the mass of MgCO3:
\[ \text{Mass of MgCO}_3 = \text{moles} \times \text{molar mass} = 0.1 \, \text{moles} \times 84 \, \text{g/mol} = 8.4 \, \text{g} \]

### Step 5: Calculate the mass of CaCO3 in the mixture.
The total mass of the mixture is given as 18.4 g. Therefore, the mass of CaCO3 can be calculated as:
\[ \text{Mass of CaCO}_3 = \text{Total mass} - \text{Mass of MgCO}_3 = 18.4 \, \text{g} - 8.4 \, \text{g} = 10.0 \, \text{g} \]

### Step 6: Calculate the moles of CaCO3.
Using the molar mass of CaCO3:
- Molar mass of CaCO3 = Atomic mass of Ca (40 g/mol) + Atomic mass of C (12 g/mol) + 3 × Atomic mass of O (16 g/mol) = 40 + 12 + 48 = 100 g/mol.

Now we can find the moles of CaCO3:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10.0 \, \text{g}}{100 \, \text{g/mol}} = 0.1 \, \text{moles} \]

### Step 7: Determine the moles of CO2 produced from CaCO3.
From the stoichiometry of the reaction, 1 mole of CaCO3 produces 2 moles of CO2. Therefore, the moles of CO2 produced from 0.1 moles of CaCO3 is:
\[ \text{Moles of CO}_2 = 2 \times \text{Moles of CaCO}_3 = 2 \times 0.1 = 0.2 \, \text{moles} \]

### Step 8: Calculate the volume of CO2 produced at STP.
At STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 liters. Therefore, the volume of CO2 produced is:
\[ \text{Volume of CO}_2 = \text{Moles of CO}_2 \times 22.4 \, \text{L/mol} = 0.2 \, \text{moles} \times 22.4 \, \text{L/mol} = 4.48 \, \text{L} \]

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

---