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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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Calculate the mass of quick lime (Cao) that can be obtained by heating 50 kg of limestone which is 95% pure.

A sample of hard water contains 96 ppm of SO_(4)^(2-) and 183 ppm of HCO_(3)^(-) , with Ca^(2+) as the only cation. How many moles of CaO will be required to remove HCO_(3)^(-) from 1000 kg of this water ? If 1000 kg of this water is treated with the amount of CaO calculated above , what will be the concentration (in ppm ) of residual Ca^(2+) ions ? (Assume CaCO_(3) to be completely insoluble in water ). If the Ca^(2+) ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH ? (One ppm means one part of the substance in one million part of water , mass/mass .)

Knowledge Check

  • 50 g limestone is heated. The quantity of quicklime produced is

    A
    56g
    B
    28g
    C
    14g
    D
    10g
  • Calcium carbonate, CaCO_(3) decomposes upon heating to calcium oxide and carbon dioxide. What mass of solid calcium carbonate is required to produce 2.40 liters of carbon dioxide measured at STP?

    A
    10.7g
    B
    21.4g
    C
    50.0g
    D
    10.6g
  • 100 g CaCO_(3) reacts with 1 litre 1 N HCl . On completion of reaction how much weight of CO_(2) will be obtain

    A
    (a)`5.5 g `
    B
    (b)` 11 g`
    C
    (c )`22 g `
    D
    (d)` 33 g`
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