Em tham khảo nha :

\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{6,2}}{{62}} = 0,1mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,2mol\\
{m_{NaOH}} = 0,2 \times 40 = 8g\\
C{\% _{NaOH}} = \dfrac{8}{{193,8 + 6,2}} \times 100\%  = 4\% \\
b)\\
{n_{CuS{O_4}}} = \dfrac{{200 \times 16\% }}{{160}} = 0,2mol\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
\dfrac{{0,2}}{1} > \dfrac{{0,2}}{2} \Rightarrow CuS{O_4}\text{ dư}\\
{n_{Cu{{(OH)}_2}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{Cu{{(OH)}_2}}} = 0,1 \times 98 = 9,8g\\
{n_{CuS{O_4}d}} = 0,2 - \dfrac{{0,2}}{2} = 0,1mol\\
{m_{CuS{O_4}d}} = 0,1 \times 160 = 16g\\
{n_{N{a_2}S{O_4}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,1mol\\
{m_{N{a_2}S{O_4}}} = 0,1 \times 142 = 14,2g\\
C{\% _{CuS{O_4}}} = \dfrac{{16}}{{200 + 200 - 9,8}} \times 100\%  = 4,1\% \\
C{\% _{N{a_2}S{O_4}}} = \dfrac{{14,2}}{{200 + 200 - 9,8}} \times 100\%  = 3,64\% 
\end{array}\)

#hlong08