Câu 10:
$n_{C_2H_2}=\frac{4,032}{22,4}=0,18 mol$
$H_1=60\%$
Bảo toàn C: $n_{C_6H_6}= \frac{1}{3}.n_{C_2H_2} . H_1\%$
$= \frac{1}{3}.0,18.60\%= 0,036 mol$
$H_2=80\%$
Bảo toàn C: $n_{C_6H_5NO_2}= n_{C_6H_6}.H_2\%$
$=0,036.80\%= 0,0288 mol$
$\Rightarrow m_{sp}= 0,0288.123= 3,5424g$
=> chọn A