\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{1,12}{22,4}=0,05mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,05 0,05 ( mol )
\(m_{Zn}=n.M=0,05.65=3,25g\)
nH2 = 1,12 :22,4=0,05(mol)
pthh : Zn+2HCl ---> ZnCl2 +H2
0,05<---------------------0,05(mol)
mZn = 0,05.65=3,25 (g)