Question

Answer 1

Ta có: $m_{C_2H_5OH}=64(g)\Rightarrow n_{C_2H_5OH}=\frac{32}{23}$

$\Rightarrow n_{CH_3COOH}=\frac{32}{25}(mol)\Rightarrow m_{CH_3COOH}=76,8(g)$

Answer 2

$V_{rượu} = 1.\dfrac{8}{100} = 0,08(lít)$

$m_{rượu} = 0,08.1000.0,8 = 64(gam)$

$n_{rượu\ pư} = \dfrac{64}{46}.92\% = 1,28(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$n_{CH_3COOH} = n_{C_2H_5OH\ pư} = 1,28(mol)$
$m = 1,28.60 = 76,8(gam)$

Đáp án A

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Answer 3

A. 76,8 gam

Answer 4

Chọn A