20gm of `CaCO_(3)` on decomposition gives `CO_(2)` at STP:

Options

A. 4.48 litre
B. 22.4 litre
C. 2.24 litre
D. 4.54 litre

Answer

To solve the problem of how much carbon dioxide (CO₂) is produced from the decomposition of 20 grams of calcium carbonate (CaCO₃) at STP, we can follow these steps:

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

### Step 2: Calculate the molar mass of calcium carbonate (CaCO₃)
To find the molar mass of CaCO₃, we sum the atomic masses of its constituent elements:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (and there are 3 oxygen atoms)

Calculating the total:
\[ \text{Molar mass of CaCO}_3 = 40 + 12 + (16 \times 3) = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 3: Calculate the number of moles of CaCO₃
Using the formula for moles:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Substituting the values:
\[ \text{Number of moles of CaCO}_3 = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ moles} \]

### Step 4: Determine the moles of CO₂ produced
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, the moles of CO₂ produced will also be:
\[ \text{Number of moles of CO}_2 = 0.2 \text{ moles} \]

### Step 5: Calculate the volume of CO₂ at STP
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as:
\[ \text{Volume of CO}_2 = \text{Number of moles} \times 22.4 \text{ L/mol} \]
Substituting the values:
\[ \text{Volume of CO}_2 = 0.2 \text{ moles} \times 22.4 \text{ L/mol} = 4.48 \text{ liters} \]

### Final Answer
The volume of carbon dioxide produced from the decomposition of 20 grams of calcium carbonate at STP is **4.48 liters**.

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