To solve the problem of finding the percentage of impurities in a 500 g impure sample of CaCO₃ that produces 70 g of CaO upon heating, we can follow these steps:
### Step 1: Write the reaction
When calcium carbonate (CaCO₃) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO₂):
\[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \]
### Step 2: Calculate molar masses
- Molar mass of CaCO₃:
\[
\text{Ca} (40 \, \text{g/mol}) + \text{C} (12 \, \text{g/mol}) + 3 \times \text{O} (16 \, \text{g/mol}) = 40 + 12 + 48 = 100 \, \text{g/mol}
\]
- Molar mass of CaO:
\[
\text{Ca} (40 \, \text{g/mol}) + \text{O} (16 \, \text{g/mol}) = 40 + 16 = 56 \, \text{g/mol}
\]
- Molar mass of CO₂:
\[
\text{C} (12 \, \text{g/mol}) + 2 \times \text{O} (16 \, \text{g/mol}) = 12 + 32 = 44 \, \text{g/mol}
\]
### Step 3: Establish the stoichiometry
From the balanced equation, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂. Therefore:
- 56 g of CaO corresponds to 44 g of CO₂.
### Step 4: Calculate the amount of CO₂ produced from 70 g of CaO
Using the ratio:
\[
\text{If } 56 \, \text{g CaO} \text{ produces } 44 \, \text{g CO}_2,
\]
\[
\text{Then } 70 \, \text{g CaO} \text{ produces } \left(\frac{44}{56} \times 70\right) \, \text{g CO}_2.
\]
Calculating this:
\[
\frac{44}{56} \times 70 = \frac{44 \times 70}{56} = \frac{3080}{56} = 55 \, \text{g CO}_2.
\]
### Step 5: Calculate the total mass of products
The total mass of products (CaO + CO₂) is:
\[
70 \, \text{g (CaO)} + 55 \, \text{g (CO}_2\text{)} = 125 \, \text{g}.
\]
### Step 6: Calculate the mass of CaCO₃ in the sample
According to the law of conservation of mass, the mass of reactants equals the mass of products. Therefore, the mass of CaCO₃ in the sample is:
\[
125 \, \text{g (CaCO}_3\text{)}.
\]
### Step 7: Calculate the mass of impurities
The mass of impurities in the sample is:
\[
\text{Total mass of sample} - \text{mass of CaCO}_3 = 500 \, \text{g} - 125 \, \text{g} = 375 \, \text{g}.
\]
### Step 8: Calculate the percentage of impurities
The percentage of impurities is given by:
\[
\text{Percentage of impurities} = \left(\frac{\text{mass of impurities}}{\text{total mass of sample}}\right) \times 100 = \left(\frac{375 \, \text{g}}{500 \, \text{g}}\right) \times 100 = 75\%.
\]
### Final Answer
The percentage of impurities in the sample is **75%**.
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