a) Zn +2HCl---.>ZnCl2 +H2
b) Ta có
n\(_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)
Theo pthh
n\(_{H2}=n_{Zn}=0,1\left(mol\right)\)=>m\(_{H2}=0,2\left(g\right)\)
V\(_{H2}=0,1.22,4=2,24\left(l\right)\)
c) Theo pthh
n\(_{ZnCl2}=n_{Zn}=0,1\left(môl\right)\)
m\(_{ZnCl2}=0,1.136=13,6\left(g\right)\)
C%=\(\frac{m_{ct}}{m_{Zn}+m_{HCl}-m_{H2}}\)
\(\frac{13,6}{100+6,5-0,2}.100\%=12.79\%\)
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