Zn + 2HCl -> ZnCl₂ + H₂ 

a, n_Zn = 13/65= 0,2(mol)

m_HCl= 109,5.20%/100%=21.9(g)

n_HCl=21,9/36,5=0,6(mol)

Theo PT n_HCl = 2n_Zn= 2.0,2= 0,4(mol)<0,6(mol)

=> HCl pư dư, Zn pư hết

Theo PT: n_H2= n_Zn =0,2(mol)

V_H2=0,2.22,4=4,48(l)

b, Theo PT: n_ZnCl2=n_Zn=0,2(mol)

m_ZnCl2= 0,2.136=27,2(g)

c, mdd sau pư= 13+109,5-0,2.2=122,1(g)

C%dd ZnCl2=27,2.100%/122,1=22,28%

n_HCl dư= 0,6-0,4=0,2(mol)

mHcl

2Na + 2H₂O ->2 NaOH + H₂

n_Na = 3,2/23=0,14(mol)

Theo PT: n_NaOH= n_Na=0,14(mol)

m_NaOH=0,14.40=5,6(g)

Theo PT: n_H2=1/2n_Na=1/2.0,14=0,07(mol)

V_H2(đktc)=0,07.22,4=1,568(l)

Đọc tên NaOH : natri hiđroxit