How many grams of CO(g) are required to produce 49.4 g of acetic acid?
回答 (1)
2016-06-13 1:16 pm
Molar mass of CO = (12.01 + 16.00) g/mol = 28.01 g/mol
Molar mass of CH₃COOH = (12.01×2 + 1.008×4 + 16.00×2) g/mol = 60.05 g/mol
CH₃OH + CO → CH₃COOH
No. of moles of CH₃COOH formed = (49.4 g) / (60.05 g/mol) = 49.4/60.05 mol
No. of moles of CO required = 49.4/60.05 mol
Mass of CO required = (49.4/60.05 mol) × (28.01 mol) = 23.0 g
Molar mass of CH₃COOH = (12.01×2 + 1.008×4 + 16.00×2) g/mol = 60.05 g/mol
CH₃OH + CO → CH₃COOH
No. of moles of CH₃COOH formed = (49.4 g) / (60.05 g/mol) = 49.4/60.05 mol
No. of moles of CO required = 49.4/60.05 mol
Mass of CO required = (49.4/60.05 mol) × (28.01 mol) = 23.0 g
收錄日期: 2021-04-18 14:59:49
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