1 gm of a mixture of calcium carbonate and magnesium carbonate gave on ignition 240 mL of carbon dioxide at S.T.P. What is the percentage composition of the mixture ?

Answer

To solve the problem, we will follow these steps:

### Step 1: Write the decomposition reactions
The decomposition reactions of calcium carbonate (CaCO3) and magnesium carbonate (MgCO3) are as follows:

1. **For Calcium Carbonate:**
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
This shows that 1 mole of CaCO3 produces 1 mole of CO2.

2. **For Magnesium Carbonate:**
\[ \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2 \]
This shows that 1 mole of MgCO3 also produces 1 mole of CO2.

### Step 2: Define variables for the masses
Let:
- \( x \) = mass of CaCO3 in grams
- \( 1 - x \) = mass of MgCO3 in grams (since the total mass of the mixture is 1 gram)

### Step 3: Calculate the number of moles
Using the molar masses:
- Molar mass of CaCO3 = 100 g/mol
- Molar mass of MgCO3 = 84 g/mol

The number of moles of each compound can be calculated as follows:
- Moles of CaCO3:
\[ \text{Moles of CaCO}_3 = \frac{x}{100} \]
- Moles of MgCO3:
\[ \text{Moles of MgCO}_3 = \frac{1 - x}{84} \]

### Step 4: Calculate the volume of CO2 produced
Since 1 mole of each carbonate produces 1 mole of CO2, the total volume of CO2 produced can be calculated using the ideal gas law at STP (1 mole of gas occupies 22.4 L or 22400 mL).

- Volume of CO2 from CaCO3:
\[ \text{Volume of CO}_2 \text{ from CaCO}_3 = \frac{x}{100} \times 22400 \]

- Volume of CO2 from MgCO3:
\[ \text{Volume of CO}_2 \text{ from MgCO}_3 = \frac{1 - x}{84} \times 22400 \]

### Step 5: Set up the equation
According to the problem, the total volume of CO2 produced is 240 mL. Therefore, we can set up the equation:
\[ \frac{x}{100} \times 22400 + \frac{1 - x}{84} \times 22400 = 240 \]

### Step 6: Simplify the equation
Dividing the entire equation by 22400 gives:
\[ \frac{x}{100} + \frac{1 - x}{84} = \frac{240}{22400} \]
Calculating the right side:
\[ \frac{240}{22400} = \frac{1}{93.33} \approx 0.0107 \]

Now, we can rewrite the equation:
\[ \frac{x}{100} + \frac{1 - x}{84} = 0.0107 \]

### Step 7: Solve for x
To solve for \( x \), we can multiply through by the least common multiple of 100 and 84, which is 8400:
\[ 84x + 100(1 - x) = 8400 \times 0.0107 \]
Calculating the right side:
\[ 8400 \times 0.0107 \approx 89.88 \]
So we have:
\[ 84x + 100 - 100x = 89.88 \]
Combining like terms:
\[ -16x + 100 = 89.88 \]
Rearranging gives:
\[ -16x = 89.88 - 100 \]
\[ -16x = -10.12 \implies x = \frac{10.12}{16} \approx 0.6325 \text{ grams} \]

### Step 8: Calculate the mass of MgCO3
Using \( x \):
\[ \text{Mass of MgCO}_3 = 1 - x = 1 - 0.6325 = 0.3675 \text{ grams} \]

### Step 9: Calculate the percentage composition
- Percentage of CaCO3:
\[ \text{Percentage of CaCO}_3 = \left(\frac{0.6325}{1}\right) \times 100 = 63.25\% \]

- Percentage of MgCO3:
\[ \text{Percentage of MgCO}_3 = \left(\frac{0.3675}{1}\right) \times 100 = 36.75\% \]

### Final Answer
The percentage composition of the mixture is approximately:
- **CaCO3: 63.25%**
- **MgCO3: 36.75%**