nCuCl2=0.5×0.2=0.1(mol)
a)CuCl2+2NaOH⇒Cu(OH)2+2NaCl (1)
0.1.........0.2...............0.1.............0.2
Cu(OH)2⇒CuO+H2O (2)
b)-Ta có:nNaOH=2nCuCl2=0.1×2=0.2(mol)
⇒M(NaOH)= $\frac{n}{V}$=$\frac{0.2}{0.3}$≈0.6(mol/l)
-Vì:nCu(OH)2=nCuCl2=0.1(mol)
⇒nCuO=nCu(OH)2=0.1(mol)
⇒mCuO=0.1×80=8(g)
c)nNaCl=nNaOH=0.2(mol)
⇒V(NaCl)=0.2×22.4=4.48(l)
⇒CM(NaCl)=$\frac{n}{V}$=$\frac{0.2}{4.48}$≈0.045(mol/l)