nZn = 13/65= 0,2 mol
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
a/ VH2 = 0,2 * 22,4 = 4,48 l
b/ mZnSO4 = 0,2 * (65 + 32 + 16*4) = 32,2 g
Zn + H2SO4\(\rightarrow\) ZnSO4 + H2 (1)
nzn=\(\dfrac{13}{65}=0,2mol\)
Theo PTHH (1): nZn=nH2=nZnSO4=0,2 mol
\(\Rightarrow\)VH2=0,2 . 22,4=4,48 lít
\(\Rightarrow\)mZnSO4=0,2 . 161=32,2g