The volume of `CO_(2)` evolved at STP on heating 50g `CaCO_(3)`

Options

A. 11.2 litre
B. 22.4 litre
C. 5.6 litre
D. 24.4 litre

Answer

To solve the problem of finding the volume of CO₂ evolved at STP when heating 50 g of CaCO₃, we will follow these steps:

### Step 1: Write the decomposition reaction of calcium carbonate
The decomposition of calcium carbonate (CaCO₃) upon heating can be represented by the following balanced chemical 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, we sum the atomic masses of its constituent elements:
- Calcium (Ca): 40 g/mol
- Carbon (C): 12 g/mol
- Oxygen (O): 16 g/mol (3 atoms of oxygen)

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

### Step 3: Determine the moles of CaCO₃ in 50 g
Using the molar mass, we can find the number of moles of CaCO₃ in 50 g:
\[ \text{Moles of CaCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \, \text{g}}{100 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 4: Use stoichiometry to find the moles of CO₂ produced
From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CO₂. Therefore, 0.5 moles of CaCO₃ will produce:
\[ \text{Moles of CO}_2 = 0.5 \, \text{mol} \]

### Step 5: Calculate the volume of CO₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as follows:
\[ \text{Volume of CO}_2 = \text{moles of CO}_2 \times 22.4 \, \text{L/mol} = 0.5 \, \text{mol} \times 22.4 \, \text{L/mol} = 11.2 \, \text{L} \]

### Final Answer
The volume of CO₂ evolved at STP when heating 50 g of CaCO₃ is:
\[ \text{Volume of CO}_2 = 11.2 \, \text{liters} \]

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