a) Phản ứng:
Zn + 2HCl\(\rightarrow\) ZnCl2 + H2
b)
Ta có: nH2=\(\frac{4,36}{22,4}\)=0,195 mol
Theo ptpu: nHCl=2nH2=0,39 mol;
\(\rightarrow\) CM HCl=\(\frac{0,39}{0,1}\)=3,9M
c) Theo ptpu: nZn=nH2=0,195 mol\(\rightarrow\) mZn=0,195.65=12,675 gam