Em tham khảo nha :
\(\begin{array}{l}
1)\\
hh:{N_2}(a\,mol);{N_2}O(b\,mol)\\
{M_A} = 36dvC\\
\Rightarrow \dfrac{{28a + 44b}}{{a + b}} = 36\\
\Rightarrow a:b = \dfrac{{44 - 36}}{{36 - 28}} = 1:1\\
\% {V_{{N_2}}} = \dfrac{1}{{1 + 1}} \times 100\% = 50\% \\
\% {V_{{N_2}O}} = 100 - 50 = 50\% \\
\% {N_2} = \dfrac{{1 \times 28}}{{1 \times 28 + 1 \times 44}} \times 100\% = 38,9\% \\
\% {N_2}O = 100 - 38,9 = 61,1\% \\
2)\\
a)\\
\% {N_2}O = \dfrac{{4,4}}{{4,4 + 9,2}} \times 100\% = 32,4\% \\
\% N{O_2} = 100 - 32,4 = 67,6\% \\
b)\\
{n_{{N_2}O}} = \dfrac{{4,4}}{{44}} = 0,1mol\\
{V_{{N_2}O}} = 0,1 \times 22,4 = 2,24l\\
{n_{N{O_2}}} = \dfrac{{9,2}}{{46}} = 0,2mol\\
{V_{N{O_2}}} = 0,2 \times 22,4 = 4,48l\\
\% {V_{{N_2}O}} = \dfrac{{2,24}}{{2,24 + 4,48}} \times 100\% = 33,3\% \\
\% {V_{N{O_2}}} = 100 - 33,3 = 66,7\% \\
c)\\
\% {n_{{N_2}O}} = \dfrac{{0,1}}{{0,1 + 0,2}} \times 100\% = 33,3\% \\
\% {n_{N{O_2}}} = 100 - 33,3 = 66,7\% \\
d)\\
{n_N} = 2{n_{{N_2}O}} + {n_{N{O_2}}} = 0,4mol\\
{m_N} = 0,4 \times 14 = 5,6g\\
{n_O} = {n_{{N_2}O}} + 2{n_{N{O_2}}} = 0,5mol\\
{m_O} = 0,5 \times 16 = 8g\\
\% N = \dfrac{{5,6}}{{5,6 + 8}} \times 100\% = 41,2\% \\
\% O = 100 - 41,2 = 58,8\% \\
e)\\
{M_{tb}} = \dfrac{{0,1 \times 44 + 0,2 \times 46}}{{0,3}} = 45,333g/mol
\end{array}\)
