a/ PTHH: \(Zn+H_2SO_4→ZnSO_4+H_2\)
b/ \(n_{Zn}=\dfrac{6,5}{65}=0,1(mol)\)
Theo pt: \(1:1:1:1\)
\(→n_{H_2}=0,1(mol)\)
\(→V_{H_2}=0,1.22,4=2,24(l)\)
c/ \(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2(mol)\)
Ta có: \(\dfrac{n_{Zn}}{1}<\dfrac{n_{H_2SO_4}}{1}\)
\(→H_2SO_4\) dư và dư \(0,2-0,1=0,1(mol)\)
\(→m_{H_2SO_4\,\,dư}=0,1.98=9,8(g)\)
Vậy \(H_2SO_4\) dư \(9,8g\)