How much amount of `CaCO_3` in gram having percentage purity 50 per cent produces 0.16 litre of` CO_2` at on heating?

Options

A. 1.4g
B. 4.2g
C. 2.8g
D. 0.7g

Explanation

To solve the problem, we need to determine how much calcium carbonate (CaCO₃) is required to produce 0.16 liters of carbon dioxide (CO₂) when heated, considering that the CaCO₃ has a purity of 50%. Here are the steps to find the solution:

### Step 1: Write the Decomposition Reaction
The decomposition of calcium carbonate can be represented by the following chemical equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Use Molar Volume at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. We need to find out how many moles of CO₂ are produced from 0.16 liters.

### Step 3: Calculate Moles of CO₂
Using the molar volume:
\[ \text{Moles of CO}_2 = \frac{\text{Volume of CO}_2}{\text{Molar Volume at STP}} = \frac{0.16 \, \text{liters}}{22.4 \, \text{liters/mole}} \approx 0.00714 \, \text{moles} \]

### Step 4: Relate Moles of CO₂ to Moles of CaCO₃
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CaCO₃ required will also be 0.00714 moles.

### Step 5: Calculate Mass of Pure CaCO₃
Next, we calculate the mass of pure CaCO₃ needed using its molar mass. The molar mass of CaCO₃ is approximately 100 g/mol.
\[ \text{Mass of CaCO}_3 = \text{Moles} \times \text{Molar Mass} = 0.00714 \, \text{moles} \times 100 \, \text{g/mol} \approx 0.714 \, \text{grams} \]

### Step 6: Account for Purity
Given that the purity of the CaCO₃ is 50%, we can set up the equation:
\[ \text{Purity} = \frac{\text{Mass of pure CaCO}_3}{\text{Total mass of CaCO}_3} \times 100 \]
Let the total mass of CaCO₃ be \(X\) grams. Therefore:
\[ 50\% = \frac{0.714 \, \text{grams}}{X} \times 100 \]
This simplifies to:
\[ 0.714 = 0.5X \]
Solving for \(X\):
\[ X = \frac{0.714}{0.5} = 1.428 \, \text{grams} \approx 1.4 \, \text{grams} \]

### Final Answer
The amount of CaCO₃ needed is approximately **1.4 grams**.

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