Answer:
a. 92.4%
Explanation:
Based on the reaction:
2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g)
To obtain the percent yield you need to obtain moles of trona and calculate thoeretical moles of Na₂CO₃, and the ratio of obtained moles / theoretical moles of Na₂CO₃ give percent yield, thus:
Moles of trona:
1.00 metric ton × (1x10³kg / 1 metric ton) × ( 1000moles /226.03 kg) = <em>4424 moles</em>
The theoretical moles of Na₂CO₃ that produce 4424 moles of trona are (Based on the reaction, 2 moles of trona produce 3 moles of Na₂CO₃):
4424 moles trona × (3 moles Na₂CO₃ / 2 moles trona) = <em>6636 moles of Na₂CO₃.</em>
The obtained moles of Na₂CO₃:
0.650 metric ton × (1x10³kg / 1 metric ton) × (1000 moles / 105.99kg) = <em>6133 moles</em>
The ratio of obtained moles / theoretical moles gives:
6133 moles / 6636 moles = 0.924 = <em>92.4%</em>
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Answer:
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Explanation:
Answer:
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Explanation:
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Answer:
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