Consider the following reaction.
3H2(g) + WO3(s) → 3H2O(g) + W(s)
When 500.0 g of WO3(s) react with excess hydrogen, 375 g of W(s) are obtained. What is the percent yield?
Consider the following reaction. 3H2(g) + WO3(s) → 3H2O(....?
2017-05-09 1:52 pm
回答 (1)
2017-05-09 5:55 pm
Molar mass of WO₃ = (183.9 + 16.0×3) g/mol = 231.9 g/mol
No. of moles of 500.0 g of WO₃ = (500.0 g) / (231.9 g/mol) = 2.156 mol
3H₂(g) + WO₃(s) → 3H₂O(g) + W(s)
Mole ratio WO₃ : W = 1 : 1
Max. no. of moles of W obtained = (2.156 mol) × 1 = 2.156 mol
Molar mass of W = 183.9 g/mol
Theoretical yield of W = (2.156 mol) × (183.9 g/mol) = 396.5 g
Percent yield = (375/396.5) × 100% = 94.6%
No. of moles of 500.0 g of WO₃ = (500.0 g) / (231.9 g/mol) = 2.156 mol
3H₂(g) + WO₃(s) → 3H₂O(g) + W(s)
Mole ratio WO₃ : W = 1 : 1
Max. no. of moles of W obtained = (2.156 mol) × 1 = 2.156 mol
Molar mass of W = 183.9 g/mol
Theoretical yield of W = (2.156 mol) × (183.9 g/mol) = 396.5 g
Percent yield = (375/396.5) × 100% = 94.6%
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