$a,PTPƯ:2Na+2_2H_5OH\xrightarrow{} 2C_2H_5ONa+H_2↑$
$b,n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$c,Theo$ $pt:$ $n_{C_2H_5ONa}=n_{C_2H_5OH}=0,2mol.$
$⇒m_{C_2H_5ONa}=0,2.68=13,6g.$
$d,PTPƯ:2Na+2CH_3COOH\xrightarrow{} 2CH_3COONa+H_2↑$
$n_{CH_3COOH}=\dfrac{100.6,9\%}{60}=0,115mol.$
$Theo$ $pt:$ $n_{CH_3COONa}=n_{Na}=n_{CH_3COOH}=0,115mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,0575mol.$
$⇒C\%_{CH_3COONa}=\dfrac{0,115.82}{(0,115.23)+100-(0,0575.2)}.100\%=9,2\%$
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