Đáp án:
$x=19,6$
Giải thích các bước giải:
$\text{Ta có sơ đồ:}$
\(\left\{ \begin{array}{l}
Cu:0,03\\
Mg:0,09
\end{array} \right. + \left\{ \begin{array}{l}
KN{O_3}:007\\
{H_2}S{O_4}:0,16
\end{array} \right. \to d{\rm{d}}\left\{ \begin{array}{l}
SO_4^{2 - }:0,16\\
C{u^{2 + }}:0,03\\
M{g^{2 + }}:0,09\\
{K^ + }:0,07\\
NH_4^ + :a
\end{array} \right. + khí\left\{ \begin{array}{l}
N:x\\
O:y
\end{array} \right. + {H_2}O\)
\(\begin{array}{l}
BTĐT:\ 2{n_{SO_4^{2 - }}} = 2{n_{C{u^{2 + }}}} + 2{n_{M{g^{2 + }}}} + {n_{{K^ + }}} + {n_{NH_4^ + }}\\
\to {n_{NH_4^ + }} = a = 0,01\ mol
\end{array}\)
\(BTNT\ N:{n_{KN{O_3}}} = {n_{NH_4^ + }} + {n_{N(khí)}} \to {n_{N(khí)}} = 0,06\ mol\)
\(BTNT\ H:2{n_{{H_2}S{O_4}}} = 4{n_{NH_4^ + }} + 2{n_{{H_2}O}} \to {n_{{H_2}O}} = 0,14\ mol\)
\(BTNTO:3{n_{KN{O_3}}} + 4{n_{{H_2}S{O_4}}} = 4{n_{SO_4^{2 - }}} + {n_{{H_2}O}} + {n_{O(khí)}}\)
\(\begin{array}{l}
\to {n_{O(khí)}} = 0,07\ mol\\
\to {m_{(khí)}} = {m_{O(khí)}} + {m_{N(khí)}} = 1,96\ g\\
\to \overline M = \dfrac{{1,96}}{{0,05}} = 39,2 \to x = d = \dfrac{{39,2}}{2} = 19,6
\end{array}\)
