a) Zn + 2HCl -> ZnCl2 + H2 (1)
b) mHCl = \(\dfrac{73.20}{100}\) = 14,6 => nHCl = \(\dfrac{14,6}{36,5}\) = 0,4(mol)
Theo PT (1) ta có: n\(ZnCl_2\) = \(\dfrac{1}{2}\)nHCl = \(\dfrac{1}{2}\).0,4 = 0,2(mol)
=> m\(ZnCl_2\) = 0,2.136 = 27,2(g)
c) Theo PT (1) ta có: n\(H_2\) = \(\dfrac{1}{2}\)nHCl = \(\dfrac{1}{2}\).0,4 = 0,2(mol)
=> V\(H_2\) = 0,2.22,4 = 4,48(l)