\(n_{SO_3}=\dfrac{100}{80}=1,25\left(mol\right)\\ n_{H_2SO_4\left(bđ\right)}=1.3=3\left(mol\right)\\ m_{H_2SO_4\left(bđ\right)}=3.98=294\left(g\right)\\ m_{ddH_2SO_4\left(bđ\right)}=1000.1.1,12=1120\left(g\right)\\ m_{ddH_2SO_4\left(sau\right)}=1120+100=1220\left(g\right)\)
PTHH: SO3 + H2O ---> H2SO4
1,25-------------->1,25
=> \(m_{H_2SO_4\left(sau\right)}=294+1,25.98=416,5\left(g\right)\)
=> \(C\%_{H_2SO_4\left(sau\right)}=\dfrac{416,5}{1220}.100\%=34,14\%\)