a) PTHH: Zn + 2HCl→ ZnCl₂ + H₂ (1)              

       \({n_{Zn}} = \frac{m}{M} = \frac{{1,3}} = 0,02(mol)\)                     

b) Theo PT (1): \({n_} = {n_{Zn}} = 0,02(mol)\)                 

                VH2=n.22,4=0,02.22,4=0,448l        

c) Theo PT (1): \({n_{HCl}} = 2{n_{Zn}} = 2.0,02 = 0,04(mol)\)         

         \({C_{{M_{HCl}}}} = \frac{n}{{{V_{dd}}}} = \frac{{0,04}}{{0,4}} = 0,1M\)

d)  nCu=mM=0,880=0,01mol

 Lập tỉ lệ: \(\frac{{0,02}}{1}\,\,\;\rangle \;\;\frac{{0,01}}{1}\)  H₂ dư, CuO phản ứng hết.       

         Theo PT (2):  nCu=nCuO=0,01(mol)

         \({m_{Cu}} = n.M = 0,01.64 = 0,64(g)\)