a) PTHH: Zn + 2HCl \(\rightarrow\) ZnCl2 + H2\(\uparrow\)
b) n\(H_2\) = \(\frac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: nZn = n\(H_2\) = 0,2 (mol)
=> mZn = 0,2.65 = 13 (g)
c) Đổi: 80ml = 0,08 (l)
Theo PT: nHCl = 2n\(H_2\) = 2.0,2 = 0,4 (mol)
=> CM HCl= \(\frac{0,4}{0,08}=5\left(M\right)\)
nH2= 4.48/22.4=0.2 mol
Zn + 2HCl --> ZnCl2 + H2
0.2___0.4___________0.2
mZn= 0.2*65=13g
CM HCl= 0.4/0.08=5M