How much amount of `CaCO_3` in gram having percentage purity 50 per cent produces 0.46 litre of` CO_2` at on heating?

Options

A. 1g
B. 2g
C. 3g
D. 4g

Answer

To solve the problem of how much amount of `CaCO3` with 50% purity produces 0.46 liters of `CO2` upon heating, we can follow these steps:

### Step 1: Write the decomposition reaction
The decomposition of calcium carbonate can be represented by the following equation:
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]

### Step 2: Use the molar volume of gas
According to Avogadro's law, at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters.

### Step 3: Calculate moles of `CO2`
We need to find out how many moles of `CO2` are produced from 0.46 liters. Using the molar volume:
\[ \text{Moles of } CO_2 = \frac{\text{Volume}}{\text{Molar Volume}} = \frac{0.46 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.0205 \, \text{moles} \]

### Step 4: Relate moles of `CaCO3` to moles of `CO2`
From the balanced equation, we see that 1 mole of `CaCO3` produces 1 mole of `CO2`. Therefore, the moles of `CaCO3` required to produce 0.0205 moles of `CO2` is also 0.0205 moles.

### Step 5: Calculate the mass of `CaCO3`
Now, we can calculate the mass of `CaCO3` needed. The molar mass of `CaCO3` is approximately 100 g/mol. Thus, the mass required is:
\[ \text{Mass of } CaCO_3 = \text{Moles} \times \text{Molar Mass} = 0.0205 \, \text{moles} \times 100 \, \text{g/mol} = 2.05 \, \text{grams} \]

### Step 6: Account for purity
Since the `CaCO3` has a purity of 50%, we need to find the total mass of the impure sample that contains 2.05 grams of pure `CaCO3`. Let the total mass of the sample be \( x \) grams. The equation for purity is:
\[ \frac{50}{100} \times x = 2.05 \]
Solving for \( x \):
\[ x = \frac{2.05 \times 100}{50} = 4.1 \, \text{grams} \]

### Final Answer
Thus, the amount of `CaCO3` with 50% purity required to produce 0.46 liters of `CO2` is approximately **4.1 grams**.

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