a, Ta có pthh
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
b,Theo đề bài ta có
nZn=\(\dfrac{6,5}{65}=0,1\left(mol\right)\)
Theo pthh
nH2=nZn=0,1 mol
\(\Rightarrow VH2_{\left(\text{đ}ktc\right)}=0,1.22,4=2,24\left(l\right)\)
c,Theo pthh
nHCl=2nZn=2.0,1=0,2 mol
\(\Rightarrow mHCl=0,2.36,5=7,3\left(g\right)\)