a)\(Zn+2HCl-->ZnCl2+H2\)
b)\(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
\(n_{H2}=n_{Zn}=0,1\left(mol\right)\)
\(V_{H2}=0,1.22,4=2,24\left(l\right)\)
c)\(n_{ZnCl2}=n_{zN}=0,1\left(mol\right)\)
\(m_{ZnCl2}=0,1.136=13,6\left(g\right)\)
Zn+2HCl-->ZnCl2+H2
0,1-------------0,1---0,1 mol
nZn=6,5\65=0,1 mol
=>VH2=0,1.22,4=2,24 l
=>mZnCl2=0,1.136=13,6 g