A 200 gm sample of `CaCO_(3)` having 40 % purity is heated. Find
moles of `CO_(2)` obtained.

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the mass of pure `CaCO3` in the sample
Given that the sample has 40% purity, we can find the mass of pure `CaCO3` in the 200 g sample.

\[ \text{Mass of pure } CaCO3 = \text{Purity} \times \text{Total mass} = \frac{40}{100} \times 200 \text{ g} = 80 \text{ g} \]

### Step 2: Calculate the molar mass of `CaCO3`
The molar mass of `CaCO3` can be calculated as follows:
- Atomic mass of Calcium (Ca) = 40 g/mol
- Atomic mass of Carbon (C) = 12 g/mol
- Atomic mass of Oxygen (O) = 16 g/mol

\[ \text{Molar mass of } CaCO3 = 40 + 12 + (16 \times 3) = 40 + 12 + 48 = 100 \text{ g/mol} \]

### Step 3: Calculate the number of moles of `CaCO3`
Now, we can calculate the number of moles of `CaCO3` using the mass of pure `CaCO3` and its molar mass.

\[ \text{Moles of } CaCO3 = \frac{\text{Mass of } CaCO3}{\text{Molar mass of } CaCO3} = \frac{80 \text{ g}}{100 \text{ g/mol}} = 0.8 \text{ moles} \]

### Step 4: Determine the moles of `CO2` produced
From the reaction of `CaCO3` when heated, we know that:

\[ CaCO3 \rightarrow CaO + CO2 \]

This indicates that 1 mole of `CaCO3` produces 1 mole of `CO2`. Therefore, if we have 0.8 moles of `CaCO3`, it will produce the same amount of `CO2`.

\[ \text{Moles of } CO2 = \text{Moles of } CaCO3 = 0.8 \text{ moles} \]

### Final Answer
The moles of `CO2` obtained from heating the 200 g sample of `CaCO3` with 40% purity is **0.8 moles**.

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