$6$ g of ethanoic acid and $2.3$ g of ethanol is mixed in the presence of conc. sulphuric acid. The number of moles of ethanoic acid and methanol used and the ester produced are respectively:

1, 5,

ethyl methanoate

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0.1, 0.05,

methyl ethanoate

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0.1, 0.05,

ethyl ethanoate

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1, 5,

methyl methanoate

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Solution

The correct option is C $0.1, 0.05,$ ethyl ethanoate
$Ethanoic acid \Rightarrow C H_{3} C O O H (M W = 60 u)$
$Ethanol \Rightarrow C_{2} H_{5} O H (M W = 32 u)$
Moles of ethanoic acid $= \frac{6}{60} = 0.1$
Moles of ethanol $= \frac{2.3}{46} = 0.05$

$C H_{3} C O O H + C_{2} H_{5} O H H^{+} - - \to C H_{3} C O O C_{2} H_{5} (ethyl ethanoate)$

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6 g of ethanoic acid and 2.3 g of ethanol is mixed in the presence of conc. sulphuric acid. The number of moles of ethanoic acid and methanol used and the ester produced are respectively:

The correct option is C 0.1,0.05, ethyl ethanoateEthanoic acid⇒CH3COOH(MW=60 u) Ethanol⇒C2H5OH(MW=32 u) Moles of ethanoic acid =660=0.1 Moles of ethanol =2.346=0.05 CH3COOH+C2H5OHH+−−→CH3COOC2H5(ethyl ethanoate)

$6$ g of ethanoic acid and $2.3$ g of ethanol is mixed in the presence of conc. sulphuric acid. The number of moles of ethanoic acid and methanol used and the ester produced are respectively: