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Limestone (CaCO(3)) decomposes into quic...

Limestone (`CaCO_(3)`) decomposes into quicklime (CaO) on strong heating. How much quantity of limestone will be required to prepare 100 kg of quicklime?

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To determine how much limestone (CaCO₃) is required to prepare 100 kg of quicklime (CaO), we can follow these steps: ### Step 1: Write the decomposition reaction The decomposition of calcium carbonate (CaCO₃) into calcium oxide (CaO) and carbon dioxide (CO₂) can be represented by the following chemical equation: \[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \] ### Step 2: Calculate the molar masses Next, we need to calculate the molar masses of the compounds involved: - Molar mass of CaCO₃: - Ca: 40 g/mol - C: 12 g/mol - O: 16 g/mol × 3 = 48 g/mol - Total: 40 + 12 + 48 = 100 g/mol - Molar mass of CaO: - Ca: 40 g/mol - O: 16 g/mol - Total: 40 + 16 = 56 g/mol ### Step 3: Determine the stoichiometry From the balanced equation, we see that 1 mole of CaCO₃ produces 1 mole of CaO. Therefore, the molar ratio is 1:1. ### Step 4: Calculate the amount of CaCO₃ needed for 100 kg of CaO We need to find out how many grams of CaCO₃ are required to produce 100 kg (or 100,000 g) of CaO. Using the molar masses calculated: - 1 mole of CaO (56 g) is produced from 1 mole of CaCO₃ (100 g). Now, we can set up a proportion to find the required mass of CaCO₃: \[ \text{Mass of CaCO}_3 = \left( \frac{\text{Mass of CaCO}_3}{\text{Mass of CaO}} \right) \times \text{Mass of CaO} \] Substituting the known values: \[ \text{Mass of CaCO}_3 = \left( \frac{100 \, \text{g}}{56 \, \text{g}} \right) \times 100,000 \, \text{g} \] ### Step 5: Perform the calculation Calculating the above expression: \[ \text{Mass of CaCO}_3 = \left( \frac{100}{56} \right) \times 100,000 \] \[ = 1.7857 \times 100,000 \approx 178,570 \, \text{g} \] ### Step 6: Convert to kg To convert grams to kilograms: \[ \text{Mass of CaCO}_3 = \frac{178,570 \, \text{g}}{1000} \approx 178.57 \, \text{kg} \] ### Final Answer Approximately **178.57 kg** of limestone (CaCO₃) is required to prepare 100 kg of quicklime (CaO). ---

To determine how much limestone (CaCO₃) is required to prepare 100 kg of quicklime (CaO), we can follow these steps: ### Step 1: Write the decomposition reaction The decomposition of calcium carbonate (CaCO₃) into calcium oxide (CaO) and carbon dioxide (CO₂) can be represented by the following chemical equation: \[ \text{CaCO}_3 (s) \rightarrow \text{CaO} (s) + \text{CO}_2 (g) \] ### Step 2: Calculate the molar masses ...
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