Đáp án:
\(\begin{array}{l}
b)\\
{m_{Cu{{(N{O_3})}_2}}} = 5,64g\\
c)\\
C{\% _{Cu{{(N{O_3})}_2}}} = 2,79\% \\
C{\% _{HN{O_3}}} = 13,45\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
CuO + 2HN{O_3} \to Cu{(N{O_3})_2} + {H_2}O\\
b)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{{2,4}}{{80}} = 0,03mol\\
{m_{HN{O_3}}} = \dfrac{{200 \times 15,5}}{{100}} = 31g\\
{n_{HN{O_3}}} = \dfrac{m}{M} = \dfrac{{31}}{{63}} = 0,492mol\\
\dfrac{{0,03}}{1} < \dfrac{{0,492}}{2} \Rightarrow HN{O_3}\text{ dư}\\
{n_{Cu{{(N{O_3})}_2}}} = {n_{CuO}} = 0,03mol\\
{m_{Cu{{(N{O_3})}_2}}} = n \times M = 0,03 \times 188 = 5,64g\\
c)\\
{m_{{\rm{dd}}spu}} = 2,4 + 200 = 202,4g\\
{n_{HN{O_3}pu}} = 2 \times {n_{CuO}} = 2 \times 0,02 = 0,06mol\\
{m_{HN{O_3}pu}} = n \times M = 0,06 \times 63 = 3,78g\\
{m_{HN{O_3}d}} = 31 - 3,78 = 27,22g\\
C{\% _{Cu{{(N{O_3})}_2}}} = \dfrac{m}{{{m_{{\rm{dd}}}}}} \times 100\% = \dfrac{{5,64}}{{202,4}} \times 100\% = 2,79\% \\
C{\% _{HN{O_3}}} = \dfrac{m}{{{m_{{\rm{dd}}}}}} \times 100\% = \dfrac{{27,22}}{{202,4}} \times 100\% = 13,45\%
\end{array}\)
