Em tham khảo nha :
\(\begin{array}{l}
3)\\
a)\\
{n_{{O_2}}} = 0,25mol\\
{V_{{O_2}}} = n \times 22,4 = 0,25 \times 22,4 = 5,6l\\
{n_{C{O_2}}} = 0,175mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,175 \times 22,4 = 3,92l\\
{n_{S{O_2}}} = 0,5mol\\
{V_{S{O_2}}} = n \times 22,4 = 0,5 \times 22,4 = 11,2l\\
b)\\
{n_{{N_2}}} = \dfrac{m}{M} = \dfrac{{2,8}}{{28}} = 0,1mol\\
{V_{{N_2}}} = n \times 22,4 = 0,1 \times 22,4 = 2,24l\\
{n_{C{O_2}}} = \dfrac{m}{M} = \dfrac{{8,8}}{{44}} = 0,2mol\\
{V_{C{O_2}}} = n \times 22,4 = 0,2 \times 22,4 = 4,48l\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{32}}{{32}} = 1mol\\
{V_{{O_2}}} = n \times 22,4 = 1 \times 22,4 = 22,4l\\
c)\\
{n_{{H_2}}} = \dfrac{{{N_{{H_2}}}}}{{{N_A}}} = \dfrac{{6 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1mol\\
{V_{{H_2}}} = n \times 22,4 = 1 \times 22,4 = 22,4l\\
{n_{N{O_2}}} = \dfrac{{{N_{N{O_2}}}}}{{{N_A}}} = \dfrac{{9 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 1,5mol\\
{V_{{H_2}}} = n \times 22,4 = 1,5 \times 22,4 = 33,6l\\
4)\\
a)\\
{M_X} = 16{M_{{H_2}}} = 16 \times 2 = 32dvC\\
\Rightarrow X:\text{Khí Oxi}({O_2})\\
b)\\
{M_A} = 14{M_{{H_2}}} = 14 \times 2 = 28dvC\\
\Rightarrow A:\text{Khí nito}({N_2})
\end{array}\)
