Đáp án:
$a,V_{H_2}=3,36l.$
$b,m_{CH_3COONa}=24,6g.$
Giải thích các bước giải:
$a,PTPƯ:2Na+2CH_3COOH\xrightarrow{} 2CH_3COONa+H_2↑$
$n_{CH_3COOH}=\dfrac{18}{60}=0,3mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$b,Theo$ $pt:$ $n_{CH_3COONa}=n_{CH_3COOH}=0,3mol.$
$⇒m_{CH_3COONa}=0,3.82=24,6g.$
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