Calculate the amount of lime that can be produced by heating 100g to 90% pure limestone

Explanation

To solve the problem of calculating the amount of lime (CaO) that can be produced by heating 100g of 90% pure limestone (CaCO3), we can follow these steps:

### Step 1: Determine the mass of pure limestone
Since the limestone is 90% pure, we first need to calculate the mass of pure CaCO3 in the 100g sample.

\[ \text{Mass of pure CaCO}_3 = 100 \, \text{g} \times 0.90 = 90 \, \text{g} \]

### Step 2: Write the chemical reaction
The decomposition of calcium carbonate (CaCO3) upon heating can be represented by the following chemical equation:

\[ \text{CaCO}_3 \, (s) \rightarrow \text{CaO} \, (s) + \text{CO}_2 \, (g) \]

### Step 3: Calculate the molar mass of CaCO3 and CaO
Next, we need the molar masses of CaCO3 and CaO to find out how much CaO can be produced from the available CaCO3.

- Molar mass of CaCO3:
- Ca: 40.08 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol × 3 = 48.00 g/mol
- Total = 40.08 + 12.01 + 48.00 = 100.09 g/mol

- Molar mass of CaO:
- Ca: 40.08 g/mol
- O: 16.00 g/mol
- Total = 40.08 + 16.00 = 56.08 g/mol

### Step 4: Calculate the amount of CaO produced
From the balanced equation, we see that 1 mole of CaCO3 produces 1 mole of CaO. Therefore, we can set up a proportion based on the molar masses:

\[ \text{Mass of CaO produced} = \left( \frac{\text{Molar mass of CaO}}{\text{Molar mass of CaCO}_3} \right) \times \text{Mass of pure CaCO}_3 \]

Substituting the values:

\[ \text{Mass of CaO produced} = \left( \frac{56.08 \, \text{g/mol}}{100.09 \, \text{g/mol}} \right) \times 90 \, \text{g} \]

Calculating this gives:

\[ \text{Mass of CaO produced} \approx 50.36 \, \text{g} \]

### Final Answer
The amount of lime (CaO) that can be produced by heating 100g of 90% pure limestone is approximately **50.36 grams**.

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