`n_(CuCl_2)=\frac{2,7}{135}=0,02(mol)`
`n_(NaOH)=\frac{30.10%}{40}=0,075(mol)`
`CuCl_2+2NaOH->Cu(OH)_2+2NaCl`
$Cu(OH)_2\xrightarrow{t^o}CuO+H_2O$
Theo PT
`n_(CuO)=n_(Cu(OH)_2)=n_(CuCl_2)=0,002(mol)`
`=>m_(CuO)=0,02.80=1,6(g)`
`b,`
Theo PT`
`n_(NaCl)=n_(NaOH (pư))=2n_(CuCl_2)=0,04(mol)`
`=>m_(NaCl)=0,004.58,5=2,34(g)`
`n_(NaOH (dư))=0,075-0,04=0,035(mol)`
`m_(NaOH)=0,035.40=1,4(g)`