Đáp án:

$\begin{gathered}   \% {m_{NaN{O_3}}} = 31,1\%  \hfill \\   \% {m_{Cu{{(N{O_3})}_2}}} = 68,9\%  \hfill \\  \end{gathered} $

Giải thích các bước giải:

 a. Phương trình hóa học:

$NaN{O_3}\xrightarrow{{{t^o}}}NaN{O_2} + \frac{1}{2}{O_2}$

$2Cu{(N{O_3})_2}\xrightarrow{{{t^o}}}2CuO + 4N{O_2} + {O_2} $

b. 

Gọi số mol của $NaNO_3$ và $Cu(NO_3)_2$ lần lượt là x và y mol

$85x + 188y = 27,3\,\,(*)$

PTHH:

$\begin{gathered} NaN{O_3}\xrightarrow{{{t^o}}}NaN{O_2} + \frac{1}{2}{O_2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,0,5x\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ 2Cu{(N{O_3})_2}\xrightarrow{{{t^o}}}2CuO + 4N{O_2} + {O_2} \hfill \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,2y\,\,\,\,\,\,\,\,\,\,\,\,\,0,5y\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $

$ \to {n_{khi}} = {n_{{O_2}}} + {n_{N{O_2}}} = 0,5x + 2y + 0,5y = \frac{{6,72}}{{22,4}} = 0,3\,\,mol\,\,\,(**)$

Từ (*) và (**) suy ra: $x = y = 0,1$

$\begin{gathered} \% {m_{NaN{O_3}}} = \frac{{0,1.85}}{{27,3}} \cdot 100\% = 31,1\% \hfill \\ \to \% {m_{Cu{{(N{O_3})}_2}}} = 100\% - 31,1\% = 68,9\% \hfill \\ \end{gathered}$