Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{n_{CuO}} = 0,25mol
\end{array}\)
Theo định luật bảo toàn nguyên tố, ta có:
\(\begin{array}{l}
{n_{Cu(banđầu)}} = {n_{CuO}} = 0,25mol\\
\to {m_{Cu(banđầu)}} = 16g\\
{N^{5 + }} + 3e \to {N^{2 + }}\\
Cu \to C{u^{2 + }} + 2e
\end{array}\)
Trong X, Cu dư => \({m_{Cu}}\)=9,6g
\(\begin{array}{l}
\to {n_{Cu}} = 0,15mol\\
\to {n_{Cu(trongCuO)}} = 0,25 - 0,15 = 0,1mol\\
\to {m_{CuO}} = 8g\\
\to \% {m_{Cu}} = \dfrac{{9,6}}{{9,6 + 8}} \times 100\% = 54,55\% \\
\to \% {m_{CuO}} = 100\% - 54,55\% = 45,45\% \\
b)\\
3Cu + 8{H^ + } + 2N{O_3}^ - \to 3C{u^{2 + }} + 2NO + 4{H_2}O(*)\\
CuO + 2{H^ + } \to C{u^{2 + }} + {H_2}O(**)\\
{H^ + } + O{H^ - } \to {H_2}O(***)\\
{n_{Cu{{(OH)}_2}}} = {n_{Cu}} = 0,25mol \to {n_{O{H^ - }}} = 0,5mol\\
\to {n_{O{H^ - }}}(***) = 0,3 \times 2 - 0,5 = 0,1mol\\
{n_{{H^ + }}}(*) = 4{n_{NO}} = 0,4mol\\
{n_{{H^ + }}}(**) = 2{n_{CuO}} = 0,2mol\\
{n_{{H^ + }}}(***) = {n_{O{H^ - }}}(***) = 0,1mol\\
\to {n_{{H^ + }}} = 0,4 + 0,2 + 0,1 = 0,7mol\\
\to {m_{HN{O_3}}} = 44,1g\\
\to C{\% _{HN{O_3}}} = \dfrac{{44,1}}{{200}} \times 100\% = 22,05\%
\end{array}\)
