To calculate the percentage yield of sulfur trioxide (\(SO_3\)) from calcium carbonate (\(CaCO_3\)), we first need to determine the theoretical yield of sulfur trioxide that could be obtained from 256 tonnes of calcium carbonate.
Assuming the reaction is as follows:
\[ CaCO_3 + H_2SO_4 \rightarrow CaSO_4 + CO_2 + H_2O \]
This reaction is essentially producing \(CaSO_4\) which, further reacting, could lead to the formation of \(SO_3\) (although the stoichiometry of \(SO_3\) production isn't provided explicitly in the equation). For the purpose of this problem, we're focusing solely on the yield calculation from available data.
Calculation Steps:
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Determine the moles of \(CaCO_3\):
- Molar mass of \(CaCO_3\) = \(40.08\) (Ca) + \(12.01\) (C) + \(3 \times 16.00\) (O) = \(100.09 , \text{g/mol}\)
- 256 tonnes = \(256,000, \text{kg} = 256,000,000, \text{g}\)
- Moles of \(CaCO_3\) = \(\frac{256,000,000 , \text{g}}{100.09 , \text{g/mol}} \approx 2,556,000 , \text{mol}\)
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Theoretical yield of \(SO_3\):
- From stoichiometry, \(1 , \text{mol} , CaCO_3\) produces \(1 , \text{mol} , SO_3\) (considering \(CaSO_4\) can lead to \(SO_3\)).
- Therefore, 2,556,000 moles of \(CaCO_3\) should theoretically yield 2,556,000 moles of \(SO_3\).
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Molar mass of \(SO_3\):
- Molar mass of \(SO_3\) = \(32.07\) (S) + \(3 \times 16.00\) (O) = \(80.07 , \text{g/mol}\)
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Convert moles of \(SO_3\) to grams:
- Theoretical mass of \(SO_3\) = \(2,556,000 , \text{mol} \times 80.07 , \text{g/mol} \approx 204,140,220 , \text{g} \)
- This is equivalent to \(204,140.22, \text{kg} \approx 204.14, \text{tonnes}\)
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Calculate the percentage yield: \[ \text{Percentage yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100 \] Given that the actual yield of \(SO_3\) is \(202 , \text{tonnes}\):
\[ \text{Percentage yield} = \left( \frac{202 , \text{tonnes}}{204.14 , \text{tonnes}} \right) \times 100 \approx 98.94% \]
Thus, the percentage yield of sulfur trioxide is approximately 98.94%.
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