$n_{Al}=\dfrac{4,59}{27}=0,17(mol)$

$\overline{M}_{spk}=16,75.2=33,5$

$\Rightarrow \dfrac{n_{NO}}{n_{N_2O}}=\dfrac{44-33,5}{33,5-30}=3$

Đặt $n_{NO}=3x; n_{N_2O}=x (mol)$

Bảo toàn e: $0,17.3=3x.3+8x$

$\Leftrightarrow x=0,03$

$V_{NO}=3x.22,4=2,016l$

$V_{N_2O}=22,4x=0,672l$

Đáp án:

\( {V_{NO}}  = 2,016{\text{ lít}}\)

\( {V_{{N_2}O}}  = 0,672{\text{ lít}}\)

Giải thích các bước giải:

Ta có:

\({M_{hh}} = 16,75{M_{{H_2}}} = 16,75.2 = 33,5\)

Áp dụng quy tắc đường chéo:

\(\begin{array}{*{20}{c}}
  {NO(30)}&{}&{10,5} \\ 
  {}&{{M_{hh}}(33,5)}&{} \\ 
  {{N_2}O(44)}&{}&{3,5} 
\end{array} \to \frac{{{n_{NO}}}}{{{n_{{N_2}O}}}} = \frac{{10,5}}{{3,5}} = 3\)

Cân bằng:

\(17Al + 66HN{O_3}\xrightarrow{{}}17Al{(N{O_3})_3} + 9NO + 3{N_2}O + 33{H_2}O\)

Ta có:

\({n_{Al}} = \frac{{4,59}}{{27}} = 0,17{\text{ mol}}\)

\( \to {n_{NO}} = \frac{9}{{17}}{n_{Al}} = 0,09{\text{ mol}} \to {{\text{n}}_{{N_2}O}} = \frac{1}{3}{n_{NO}} = 0,03{\text{ mol}}\)

\( \to {V_{NO}} = 0,09.22,4 = 2,016{\text{ lít}}\)

\( \to {V_{{N_2}O}} = 0,03.22,4 = 0,672{\text{ lít}}\)