If you used 6.00g sodium bicarbonate how many moles of acetic acid could be fully reacted?
回答 (1)
2016-05-11 6:50 am
Molar mass of NaHCO₃ = 22.99 + 1.01 + 12.01 + 16.00× 3 = 84.01 g/mol
NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
n(NaHCO₃) : n(CH₃COOH) = 1 : 1
n(NaHCO₃) = (6.00 g) / (84.01 g/mol) = 0.0714 mol
n(CH₃COOH) = 0.0714 mol
NaHCO₃ + CH₃COOH → CH₃COONa + H₂O + CO₂
n(NaHCO₃) : n(CH₃COOH) = 1 : 1
n(NaHCO₃) = (6.00 g) / (84.01 g/mol) = 0.0714 mol
n(CH₃COOH) = 0.0714 mol
收錄日期: 2021-04-18 14:48:26
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