Question

1.有2.7G纯铝和100ML2.0mol/L的HCL反应后,能得到标准状况下的H2多少升?

Answer 1

1.n(Al)=(2.7g)÷(27g/mol)=0.1mol; n(HCl)=(0.1mL)*(2.0mol/L)=0.2mol

2 Al + 6 HCl = 2 AlCl3 +3 H2

所以Al过量,按HCl完全反应计算.n(H2)=0.5n(HCl)=0.1mol

V(H2)=22.4×0.1=2.24L

2.Cu(2+) + 3 I(-) = CuI + I2； I2 + 2 Na2S2O3 = 2 NaI + Na2S4O6

所以n(Cu)=n(I2)=0.5*n(Na2S2O3)=0.5*0.1000*0.02020=0.00101mol

m(Cu)=63.5*0.00101=0.0641g

含量0.0641/0.2000=32.1%