a) \(C_6H_6+HNO_3\xrightarrow[]{H_2SO_4,t^o}C_6H_5NO_2+H_2O\)
b) Ta có : \(n_{C6H6}=\dfrac{11,7}{78}=0,15\left(mol\right)\)
Theo pt : \(n_{C6H6}=n_{C6H5NO2}=0,15\left(mol\right)\)
\(\rightarrow m_{C6H5NO2\left(lt\right)}=0,15.123=18,45\left(g\right)\)
\(\rightarrow m_{C6H5NO2\left(tt\right)}=18,45.100\%=18,45\left(g\right)\)