$C_6H_5CH_3+3HONO_2 \xrightarrow{{H_2SO_4 đ, t^o}} C_6H_2(NO_2)_3CH_3+ 3H_2O$
a,
$n_{C_6H_5CH_3}=\dfrac{92}{92}=1(kmol)$
$H=90\%$ nên có $1.90\%=0,9(kmol)$ toluen phản ứng.
$\to n_{TNT}=n_{\text{toluen phản ứng}}=0,9(kmol)$
$\to m_{TNT}=0,9.(92-3+46.3)=204,3(kg)$
b,
$n_{HNO_3}=3n_{\text{toluen phản ứng}}=2,7(kmol)$
$\to m_{HNO_3}=2,7.63=170,1(kg)$