The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _______ kg. (Nearest integer)
Given: Molar mass (in g mol$^{-1}$) of Ca-40, O-16, C-12
Given: Molar mass (in g mol$^{-1}$) of Ca-40, O-16, C-12
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Correct Answer: 63
Approach Solution - 1
To determine the amount of calcium oxide (CaO) produced from heating 150 kg of limestone (CaCO₃), knowing the limestone is 75% pure, we follow these steps:
1. Calculate the mass of pure CaCO₃: The limestone is 75% pure, so the mass of pure CaCO₃ is 75% of 150 kg.
Mass of pure CaCO₃ = 0.75 × 150 kg = 112.5 kg.
2. Use the chemical reaction:
CaCO₃ (s) → CaO (s) + CO₂ (g)
3. Calculate the molar masses:
Ca = 40, C = 12, O = 16 (given).
Molar mass of CaCO₃ = 40 + 12 + (16×3) = 100 g/mol.
Molar mass of CaO = 40 + 16 = 56 g/mol.
4. Convert mass of CaCO₃ to moles:
Moles of CaCO₃ = \(\frac{112,500\, \text{g}}{100\, \text{g/mol}}\) = 1125 moles.
5. Moles of CaCO₃ to moles of CaO: The reaction shows a 1:1 molar ratio, so moles of CaO = 1125 moles.
6. Calculate the mass of CaO:
Mass of CaO = moles of CaO × molar mass of CaO = 1125 moles × 56 g/mol = 63000 g = 63 kg.
7. Verify the range: The computed mass of CaO is 63 kg, which fits perfectly within the given range of 63-63 kg.
Therefore, the amount of calcium oxide produced is 63 kg.
Approach Solution -2
Given that: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \] We start by calculating the mass of CaCO$_3$: \[ \text{mass of CaCO}_3 = \frac{150 \times 75}{100} = 112.5 \, \text{kg} \] Next, calculate the moles of CaCO$_3$: \[ n_{\text{CaCO}_3} = \frac{\text{mass}}{\text{molar mass of CaCO}_3} = \frac{1125000}{100} = 1125 \, \text{moles} \] Since each mole of CaCO$_3$ produces 1 mole of CaO, the moles of CaO formed will be the same: \[ n_{\text{CaO}} = 1125 \, \text{moles} \] Now, we calculate the mass of CaO: \[ \text{mass of CaO} = n_{\text{CaO}} \times \text{molar mass of CaO} = 1125 \times 56 = 63000 \, \text{grams} = 63 \, \text{kg} \]
Thus, the amount of calcium oxide produced is 63 kg.
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