Giải thích các bước giải:

Câu 3.

1. 

Giả sử hỗn hợp khí A nặng 100 gam

\[\begin{array}{*{20}{l}}
{ \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{{m_{C{O_2}}} = 100 \cdot 24,2\%  = 24,2\;{\rm{gam}}}\\
{{m_{S{O_2}}} = 100 \cdot 48\%  = 48\;{\rm{gam}}}\\
{{m_{{O_2}}} = 100 \cdot 15,2\%  = 15,2\;{\rm{gam}}}\\
{{m_{{N_2}}} = 100 \cdot 12,6\%  = 12,6\;{\rm{gam}}}
\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}
{{n_{C{O_2}}} = \frac{{24,2}}{{44}} = 0,55\;{\rm{mol}}}\\
{{n_{S{O_2}}} = \frac{{48}}{{64}} = 0,75\;{\rm{mol}}}\\
{{n_{{O_2}}} = \frac{{15,2}}{{32}} = 0,475\;{\rm{mol}}}\\
{{n_{{N_2}}} = \frac{{12,6}}{{28}} = 0,45\;{\rm{mol}}}
\end{array}} \right.}\\
{ \Rightarrow {n_A} = 0,55 + 0,75 + 0,475 + 0,45 = 2,225\;{\rm{mol}}}\\
{\% {V_{C{O_2}}} = \frac{{0,55}}{{2,225}} \cdot 100\%  = 24,72\% }\\
{\% {V_{S{O_2}}} = \frac{{0,75}}{{2,225}} \cdot 100\%  = 33,71\% }\\
{\% {V_{{O_2}}} = \frac{{0,475}}{{2,225}} \cdot 100\%  = 21,35\% }\\
{\% {V_{{N_2}}} = 100\%  - 24,72\%  - 33,71\%  - 21,35\%  = 20,22\% }\\
{b.}\\
{\overline {{M_A}}  = \frac{{100}}{{2,225}} \approx {\rm{44}},{\rm{94}}\;{\rm{gam/mol}}}\\
{ \Rightarrow {d_{A/kk}} = \frac{{44,94}}{{29}} = 1,55}
\end{array}\]

Câu 4.

Giả sử có 1 mol X

Gọi số mol \(N_2, H_2, NH_3\) lần lượt là \(a,b,c\)

\[ \Rightarrow {n_X} = a + b + c = 1\;{\rm{mol}}\ (1)\]

\[\begin{array}{l} {d_{X/{H_2}}} = 6,8\\ \Leftrightarrow \overline {{M_X}} = 2 \cdot 6,8 = 13,6\;{\rm{gam/mol}}\\ \Rightarrow \frac{{28a + 2b + 17c}}{1} = 13,6 \Leftrightarrow 28a + 2b + 17c = 13,6\;(2)\\ b = 3a \Leftrightarrow 3a - b = 0\;(3)\\ (1),(2),(3) \Rightarrow \left\{ {\begin{array}{*{20}{l}} {a = 0,1}\\ {b = 0,3}\\ {c = 0,6} \end{array}} \right.\\ \Rightarrow \% {V_{{N_2}}} = \frac{{0,1}}{1} \cdot 100\% = 10\% \\ \% {V_{{H_2}}} = \frac{{0,3}}{1} \cdot 100\% = 30\% \\ \% {V_{N{H_3}}} = 100\% - \left( {10\% + 30\% } \right) = 60\% \end{array}\]