a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
Theo PT: \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,3\left(mol\right)\)
\(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
⇒ mZn = 0,3.65 = 19,5 (g)
mHCl = 0,6.36,5 = 21,9 (g)
c, mZnCl2 = 0,3.136 = 40,8 (g)