.............................................................

$\rm a)\\n_{Na_2O}=\dfrac{6,2}{62}=0,1(mol)\\Na_2O+H_2O\xrightarrow[]{}2NaOH\\0,1\xrightarrow[]{}0,1\xrightarrow[]{}0,2(mol)\\\xrightarrow[]{}C\%dd_{NaOH}=\dfrac{0,2.40}{6,2+193,8}.100\%=4\%\\b)\\n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2(mol)\\2NaOH+CuSO_4\xrightarrow[]{}Na_2SO_4+Cu(OH)_2\\Ta Có:(n_{NaOH}:2)<(n_{CuSO_4}:1)\xrightarrow[]{}CuSO_4,dư\\2NaOH+CuSO_4\xrightarrow[]{}Na_2SO_4+Cu(OH)_2\\0,2\xrightarrow[]{}0,1\xrightarrow[]{}0,1\xrightarrow[]{}0,1(mol)\\\xrightarrow[]{}a=0,1.98=9,8(g)\\c)\\Cu(OH)_2\xrightarrow[]{t^o}H_2O+CuO\\0,1\xrightarrow[]{}0,1\xrightarrow[]{}0,1(mol)\\CuO+2HCl\xrightarrow[]{}CuCl_2+H_2O\\0,1\xrightarrow[]{}0,2(mol)\\\xrightarrow[]{}V_{dd_{HCl}}=\dfrac{0,2}{2}=0,1(l)$

`a)`

`Na_2O+H_2O->2NaOH`

`n_{Na_2O}={6,2}/{62}=0,1(mol)`

Theo PT: `n_{NaOH}=2n_{Na_2O}=0,2(mol)`

`->C\%_{NaOH}={0,2.40}/{6,2+193,8}.100\%=4\%`

`b)`

`n_{CuSO_4}={200.16\%}/{160}=0,2(mol)`

`2NaOH+CuSO_4->Cu(OH)_2+Na_2SO_4`

Do `{0,2}/2<0,2->CuSO_4` dư

Theo PT: `n_{Cu(OH)_2}=1/2n_{NaOH}=0,1(mol)`

`->a=0,1.98=9,8(g)`

`c)`

`Cu(OH)_2`  $\xrightarrow{t^o}$  `CuO+H_2O`

`CuO+2HCl->CuCl_2+H_2O`

Theo PT: `n_{HCl}=2n_{Cu(OH)_2}=0,2(mol)`

`->V_{dd\ HCl}={0,2}/2=0,1(l)=100(ml)`