How much amount of `CaCO_3` in gram having percentage purity 50 per cent produces 0.66 litre of` CO_2` at on heating?

Options

A. 2.94g
B. 5.88g
C. 1.45g
D. 3.42g

Explanation

To solve the problem of how much amount of `CaCO3` with a percentage purity of 50% produces 0.66 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 chemical equation:
\[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \]

### Step 2: Use Avogadro's Law
According to Avogadro's law, 1 mole of any gas occupies 22.4 liters at standard temperature and pressure (STP).

### Step 3: Calculate moles of `CO2`
First, we need to calculate the number of moles of `CO2` produced from the volume given:
\[ \text{Volume of } CO_2 = 0.66 \text{ liters} \]

Using the molar volume:
\[ \text{Moles of } CO_2 = \frac{\text{Volume of } CO_2}{\text{Molar Volume}} = \frac{0.66 \text{ L}}{22.4 \text{ L/mol}} \]

### Step 4: Calculate the moles of `CaCO3` needed
From the balanced equation, we see that 1 mole of `CaCO3` produces 1 mole of `CO2`. Therefore, the moles of `CaCO3` required is the same as the moles of `CO2` produced.

### Step 5: Calculate the mass of `CaCO3`
Next, we need to find the mass of `CaCO3` that corresponds to the moles calculated. The molar mass of `CaCO3` is approximately 100 g/mol. Thus, the mass of `CaCO3` can be calculated as:
\[ \text{Mass of } CaCO_3 = \text{Moles of } CaCO_3 \times \text{Molar Mass of } CaCO_3 \]

### Step 6: Adjust for percentage purity
Since the purity of the `CaCO3` is 50%, we need to find the total mass of `CaCO3` required to yield the calculated mass. Let \( X \) be the total mass of `CaCO3` needed. The pure `CaCO3` in \( X \) grams is:
\[ \text{Pure } CaCO_3 = \frac{50}{100} \times X = 2.94 \text{ grams} \]

### Step 7: Solve for \( X \)
Now, we can solve for \( X \):
\[ X = \frac{2.94 \text{ grams} \times 100}{50} = 5.88 \text{ grams} \]

Thus, the amount of `CaCO3` required with 50% purity to produce 0.66 liters of `CO2` is **5.88 grams**.

### Final Answer
The amount of `CaCO3` needed is **5.88 grams**.

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