\(PTHH:Zn+2HCl->ZnCl_2+H_2\)
0,05--->0,1----->0,05----->0,05
\(n_{Zn}=\dfrac{m}{M}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(m_{ZnCl_2}=n\cdot M=0,05\cdot\left(65+35,5\cdot2\right)=6,8\left(g\right)\)
\(V_{H_2\left(dktc\right)}=n\cdot22,4=0,05\cdot22,4=1,12\left(l\right)\)