Đáp án:
\(\begin{array}{l}
a)\\
{C_\% }NaOH = 4\% \\
b)\\
x = 8g\\
c)\\
{V_{{\rm{dd}}HCl}} = 0,1l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{n_{N{a_2}O}} = \dfrac{{6,2}}{{62}} = 0,1\,mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,2\,mol\\
{m_{{\rm{dd}}A}} = 6,2 + 193,8 = 200g\\
{C_\% }NaOH = \dfrac{{0,2 \times 40}}{{200}} \times 100\% = 4\% \\
b)\\
CuS{O_4} + 2NaOH \to Cu{(OH)_2} + N{a_2}S{O_4}\\
{n_{CuS{O_4}}} = \dfrac{{200 \times 16\% }}{{160}} = 0,2\,mol\\
\dfrac{{0,2}}{1} > \dfrac{{0,2}}{2} \Rightarrow\text{ $CuSO_4$ dư} \\
{n_{Cu{{(OH)}_2}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
Cu{(OH)_2} \to CuO + {H_2}O\\
{n_{CuO}} = {n_{Cu{{(OH)}_2}}} = 0,1\,mol\\
x = {m_{CuO}} = 0,1 \times 80 = 8g\\
c)\\
CuO + 2HCl \to CuC{l_2} + {H_2}O\\
{n_{HCl}} = 2{n_{CuO}} = 0,2\,mol\\
{V_{{\rm{dd}}HCl}} = \dfrac{{0,2}}{2} = 0,1l
\end{array}\)
