$n_{O2}$=$\frac{m}{M}$=$\frac{9.6}{32}$=0
e(mol)
ptpư
$C_{2}$$H_{4}$+3$O_{2}$→2$CO_{2}$+2$H_{2}$$O$
a.từ pt suy ra
$n_{C2H4}$=$\frac{1}{3}$*$n_{O2}$=0.1
$m_{C2H4}$=n*M=0.1*28=2.8(g)
b.$n_{C2H5OH}$=$\frac{m}{M}$=$\frac{3.6}{46}$=$\frac{9}{115}$(mol)
ptpư:
$C_{2}$$H_{4}$+$H_{2}$$O$→$C_{2}$$H_{5}$$OH$
từ pt suy ra:
$n_{C2H4pư}$=$n_{C2H5OH}$=$\frac{9}{115}$(mol)
H=$\frac{npư}{nthamgia}$ *100=$\frac{9/115}{0.1}$ *100=78.26%
