2Al+2OH-+2H2O=2AlO2-+3H2↑…③
由題可知n(H2)=
0.672L |
22.4L/mol |
則反應(yīng)③消耗的n(NaOH)=0.0200mol,n(Al)=0.0200mol,
反應(yīng)②消耗的n(NaOH)=2.00mol?L-1×0.210L-0.0200mol=0.400mol,
則反應(yīng)①消耗的n(Al)=0.100mol,n(Cu2+)=0.150mol
故加入鋁粉的質(zhì)量為 m(Al)=(0.100mol+0.0200mol)×27.0g?mol-1=3.24g,
硫酸銅溶液的濃度為c(CuSO4)=
0.15mol |
0.03L |
答:(1)加入鋁粉的質(zhì)量為3.24g;
(2)原硫酸銅溶液的物質(zhì)的量濃度為5mol/L.