Đáp án:
\(\begin{array}{l}
2)\\
pH = 13\\
3)\\
a)\\
{m_{Cu}} = 9,6g\\
{m_{CuO}} = 8g\\
b)\\
m = 54,4g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2)\\
{n_{HCl}} = 0,1 \times 0,2 = 0,02mol\\
\Rightarrow {n_{{H^ + }}} = {n_{HCl}} = 0,02mol\\
{n_{KOH}} = 0,15 \times 0,3 = 0,045mol\\
\Rightarrow {n_{O{H^ - }}} = {n_{KOH}} = 0,045mol\\
{H^ + } + O{H^ - } \to {H_2}O\\
\dfrac{{0,02}}{1} < \dfrac{{0,045}}{1} \Rightarrow O{H^ - }\text{ dư}\\
{n_{O{H^ - }d}} = 0,045 - 0,02 = 0,025mol\\
[O{H^ - }] = \dfrac{{0,025}}{{0,1 + 0,15}} = 0,1M\\
pOH = - \log (0,1) = 1\\
pH = 14 - 1 = 13\\
3)\\
a)\\
3Cu + 8HN{O_3} \to 3Cu{(N{O_3})_2} + 2NO + 4{H_2}O\\
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
{n_{NO}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
\Rightarrow {n_{Cu}} = \dfrac{3}{2}{n_{NO}} = \dfrac{3}{2} \times 0,1 = 0,15mol\\
{m_{Cu}} = 64 \times 0,15 = 9,6g\\
\Rightarrow {m_{CuO}} = 17,6 - 9,6 = 8g\\
b)\\
3Mg + 8HN{O_3} \to 3Mg{(N{O_3})_2} + 2NO + 4{H_2}O\\
{n_{NO}} = \dfrac{{3,36}}{{22,4}} = 0,15mol\\
\Rightarrow \dfrac{2}{3}{n_{Cu}} + \dfrac{2}{3}{n_{Mg}} = 0,15mol\\
\Rightarrow {n_{Mg}} = \dfrac{{0,15 - 0,75 \times 0,15}}{{0,75}} = 0,05mol\\
{n_{CuO}} = \dfrac{8}{{80}} = 0,1mol\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{Cu}} + {n_{CuO}} = 0,15 + 0,1 = 0,25mol\\
{m_{Cu{{(N{O_3})}_2}}} = 0,25 \times 188 = 47g\\
{n_{Mg{{(N{O_3})}_2}}} = {n_{Mg}} = 0,05mol\\
{m_{Mg{{(N{O_3})}_2}}} = 0,05 \times 148 = 7,4g\\
m = 47 + 7,4 = 54,4g
\end{array}\)
