$Na_2CO_3 + 2CH_3COOH \to 2CH_3COONa + CO_2 + H_2O$
$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
Theo PTHH :
$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$
$n_{CH_3COOH} = 2n_{CO_2} = 0,2(mol)$
$m_{CH_3COOH} = 0,2.60 = 12(gam)$
$2CH_3COOH + 2Na \to 2CH_3COONa + H_2$
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$
Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} + \dfrac{1}{2}n_{C_2H_5OH} = 0,2(mol)$
$\Rightarrow n_{C_2H_5OH} = 0,2(mol)$
$m_{C_2H_5OH} = 0,2.46 = 9,2(gam)$
$\%m_{CH_3COOH} = \dfrac{12}{12 + 9,2}.100\% = 56,6\%$
$\%m_{C_2H_5OH} = 100\% - 56,6\% = 43,4\%$
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