Đáp án:

\(\begin{array}{l}
a)\\
C{\% _{{H_2}S{O_4}}} = 9,8\% \\
b)\\
{C_{{M_{NaOH}}}} = 4M\\
{m_{Cu{{(OH)}_2}}} = 19,6g
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
CuO + {H_2}S{O_4} \to CuS{O_4} + {H_2}O\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{16}}{{80}} = 0,2mol\\
{n_{{H_2}S{O_4}}} = {n_{CuO}} = 0,2mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,2 \times 98 = 19,6g\\
C{\% _{{H_2}S{O_4}}} = \dfrac{{19,6}}{{200}} \times 100\%  = 9,8\% \\
b)\\
CuS{O_4} + 2NaOH \to N{a_2}S{O_4} + Cu{(OH)_2}\\
{n_{CuS{O_4}}} = {n_{CuO}} = 0,2mol\\
{n_{NaOH}} = 2{n_{CuS{O_4}}} = 0,4mol\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,4}}{{0,1}} = 4M\\
{n_{Cu{{(OH)}_2}}} = {n_{CuS{O_4}}} = 0,2mol\\
{m_{Cu{{(OH)}_2}}} = n \times M = 0,2 \times 98 = 19,6g
\end{array}\)

`a)`

$n_{CuO}$ `=` `16/80``=0,2(mol)`

`CuO` `+` $H_{2}$$SO_{4}$ `→` $CuSO_{4}$ `+`$H_{2}O$ 

`0,2(mol)`→`0,2(mol)`→`0,2(mol)`→`0,2(mol)`

`⇒`$m_{H_{2}SO_{4}}$ `=0,2.98=19,6(g)`

`⇒` `C%`$_{H_{2}SO_{4}}$ `=` `(19,6.100)/200` `=9,8%`

`b)`

$CuSO_{4}$ `+` `2NaOH` `→` $Na_{2}$$SO_{4}$ `+`  $Cu(OH)_{2}$ 

`0,2(mol)`→`0,2(mol)`→`0,2(mol)`→`0,2(mol)`

`⇒`$C_{MNaOH}$ `=` `(0,4)/(0,1)` `=4M`

`⇒` $m_{Cu(OH)_{2}}$ `=` `0,2.98=19,2(g)`

`cuthilien`