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What is the mass of pure ethanoic acid required to neutralize 280 mL of 0.5 molar pure lime water completely? (1) 60.4 g (2) 30.2 g (3) 16.8 g (4) 8.4 g |
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Answer» (3) 16.8 g 2CH3 COOH + Ca (OH)2 → Ca (CH3 COOH)2 + 2 H2O 280 mL, 0.5 = 140 x 10–3 moles 1 mole required 2 moles = CH3COOH 140 x 10-3 required 140 x 10-3 x 2 moles CH3COOH = 0.280 mole i.e 0.280 x 60 = 16.8 g |
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