Coi như p/ứ vừa đủ
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{19,5}{65}=0,3 \left(mol\right)=n_{H_2SO_4}=n_{ZnSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4}=\dfrac{0,3\cdot98}{200}=14,7\%\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\m_{ZnSO_4}=0,3\cdot161=48,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}=218,9\left(g\right)\)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{48,3}{218,9}\cdot100\%\approx22,06\%\)