What is the mass of pure ethanoic acid required to neutralise 280 ml of 0.5molar pure lime water answer?
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We begin by writing a balanced equation for the reaction :
2 CH₃ COOH (aq) + Ca(OH)₂ (aq) —> Ca(CH₃COO)₂ (aq) + H₂O (l)
From the equation, the mole ratio is : 2 : 1
Meaning for every 2 moles of ethanoic acid we need 1 mole of lime water (Calcium hydroxide) for complete reaction.
Moles of lime water in 280 ml
(280/1000) × 0.5 = 0.14 moles.
Since the mole ratio is 2:1
Moles of ethanoic acid is :
2 × 0.14 = 0.28 moles
Molar mass of ethanoic acid :
12 + 3 +12 +16 + 16 +16 + 1 = 60g / mol
Mass of ethanoic acid is :
60 × 0.28 = 16.8 g
2 CH₃ COOH (aq) + Ca(OH)₂ (aq) —> Ca(CH₃COO)₂ (aq) + H₂O (l)
From the equation, the mole ratio is : 2 : 1
Meaning for every 2 moles of ethanoic acid we need 1 mole of lime water (Calcium hydroxide) for complete reaction.
Moles of lime water in 280 ml
(280/1000) × 0.5 = 0.14 moles.
Since the mole ratio is 2:1
Moles of ethanoic acid is :
2 × 0.14 = 0.28 moles
Molar mass of ethanoic acid :
12 + 3 +12 +16 + 16 +16 + 1 = 60g / mol
Mass of ethanoic acid is :
60 × 0.28 = 16.8 g
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