Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
______0,1__________0,1____0,1 (mol)
b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, Ta có: m dd sau pư = 6,5 + 200 - 0,1.2 = 206,3 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{0,1.136}{206,3}.100\%\approx6,59\%\)