\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Ban đầu :0,1_____0,2________________
Phứng :0,1_______0,1_______________
Sau phứng :0____0,1_________0,1________0,1
\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(n_{H2SO4}=\frac{19,6}{98}=0,2\left(mol\right)\)
\(V_{H2}=0,1.22,4=2,24\left(l\right)\)
Sau phản ứng H2SO4 còn dư
\(m_{H2SO4_{du}}=0,1.98=9,8\left(g\right)\)