Bài 11:
a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,3}{2}\) => Zn dư, HCl hết
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,15<--0,3---------->0,15
=>mZn(dư) = (0,2-0,15).65 = 3,25 (g)
b) VH2 = 0,15.24,79 = 3,7185 (l)