Đáp án:

a. $\% {m_{Cu}} = 96\% \\ \% {m_{CuO}} =  4\% $

b. ${C_{M\,\,HN{O_3}}}  = 0,18M$

Giải thích các bước giải:

${n_{NO}} = \frac{{6,72}}{{22,4}} = 0,3\,\,mol$

Sơ đồ phản ứng:

$\left\{ \begin{array}{l} \mathop {Cu}\limits^0 \\ \mathop {Cu}\limits^{ + 2} O \end{array} \right. + H\mathop N\limits^{ + 5} {O_3} \to \mathop {Cu}\limits^{ + 2} {(N{O_3})_2} + \mathop N\limits^{ + 2} O + {H_2}O$

Bảo toàn electron: $2{n_{Cu}} = 3{n_{NO}} \to {n_{Cu}} = \frac{{3.0,3}}{2} = 0,45\,\,mol$

$\begin{array}{l} {m_{Cu}} = 0,45.64 = 28,8\,\,gam\\ \to \% {m_{Cu}} = \frac{{28,8}}{{30}} \cdot 100\% = 96\% \\ \to \% {m_{CuO}} = 100\% - 96\% = 4\% \end{array}$

b. Ta có:

${m_{CuO}} = 30 - 28,8 = 1,2\,\,gam \to {n_{CuO}} = \frac{{1,2}}{{80}} = 0,015\,\,mol$

Bảo toàn nguyên tố Cu:

$n_{Cu(NO_3)_2} =n_{Cu} + n_{CuO} = 0,45 + 0,015 = 0,465$ mol

$\to {C_{M\,\,Cu{{(N{O_3})}_2}}} = \frac{{0,465}}{{1,5}} = 0,31M$

Mặt khác: ${n_{HN{O_3}\,\,ban\,\,dau}} = 1,5.1 = 1,5\,\,mol$

Bảo toàn nguyên tố N:

$\begin{array}{l} {n_{HN{O_3}\,\,phan\,\,ung}} = 2{n_{Cu{{(N{O_3})}_2}}} + {n_{NO}} = 2.0,465 + 0,3 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,= 1,23\,\,mol\\ \to {n_{HN{O_3}\,\,du}} = 1,5 - 1,23 = 0,27\,\,mol\\ \to {C_{M\,\,HN{O_3}}} = \frac{{0,27}}{{1,5}} = 0,18M \end{array}$