Zn+2HCl->ZnCl2+H2
0,1--0,2----0,1-----0,1
n Zn=0,1 mol
=>m ZnCl2=0,1.136=13,6g
=>CMHCl=\(\dfrac{0,2}{0,3}=0,67M\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1 0,1
\(m_{ZnCl_2}=136.0,1=13,6g\\
V_{H_2}=0,1.22,4=2,24l\\
C_{M\left(HCl\right)}=\dfrac{0,2}{0,3}=0,6M\)