Using the equation 2KOH = K2O + H2O, what is the amount of water in grams formed if 45.6 grams KOH are used?
回答 (1)
2017-11-07 7:08 pm
Molar mass of KOH = (39.0 + 16.0 + 1.0) g/mol = 56.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
2KOH → K₂O + H₂O
Mole ratio KOH : H₂O = 2 : 1
No. of moles of KOH used = (45.6 g) / (56.0 g/mol) = 0.814 mol
No. of moles of H₂O formed = (0.814 mol) × (1/2) = 0.407 mol
Mass of H₂O formed = (0.407 mol) × (18.0 g/mol) = 7.33 g
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
2KOH → K₂O + H₂O
Mole ratio KOH : H₂O = 2 : 1
No. of moles of KOH used = (45.6 g) / (56.0 g/mol) = 0.814 mol
No. of moles of H₂O formed = (0.814 mol) × (1/2) = 0.407 mol
Mass of H₂O formed = (0.407 mol) × (18.0 g/mol) = 7.33 g
收錄日期: 2021-04-18 17:56:10
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