(Dữ kiện liên quan đến 14,8g hỗn hợp thừa nhé)
a,
$2CH_3COOH+Zn\to (CH_3COO)_2Zn+H_2$
$Na_2CO_3+2CH_3COOH\to 2CH_3COONa+CO_2+H_2O$
$NaHCO_3+CH_3COOH\to CH_3COONa+H_2O+CO_2$
b,
$n_{H_2}=\frac{0,28}{22,4}=0,0125 mol$
$\Rightarrow n_{CH_3COOH}=0,125.2=0,025 mol$
$C_{M_{CH_3COOH}}=\frac{0,025}{0,05}=0,5M$