Đáp án:

a) \(\% {V_{C{O_2}}} = \% {V_{{O_2}}} = 46,67\% ;\% {V_{{N_2}}} = 6,66\% \)

b) c) 

\(\overline {{M_A}}  = 37,33;{d_{A/{H_2}}} = 18,665\)

Giải thích các bước giải:

 Giả sử có 100 gam hỗn hợp A

\( \to {m_{C{O_2}}} = 100.55\%  = 55{\text{ gam; }}{{\text{m}}_{{O_2}}} = 100.40\%  = 40{\text{ gam}} \to {{\text{m}}_{{N_2}}} = 100 - 55 - 40 = 5{\text{ gam}}\)

\( \to {n_{C{O_2}}} = \frac{{55}}{{44}} = 1,25{\text{ mol; }}{{\text{n}}_{{O_2}}} = \frac{{40}}{{32}} = 1,25{\text{ mol; }}{{\text{n}}_{{N_2}}} = \frac{5}{{28}} = 0,1786{\text{ mol}}\)

\({n_A} = 1,25 + 1,25 + 0,1786 = 2,6786{\text{ mol}}\)

Vì % số mol=% thể tích

\( \to \% {V_{C{O_2}}} = \% {V_{{O_2}}} = \frac{{1,25}}{{2,6786}} = 46,67\%  \to \% {V_{{N_2}}} = 100\%  - 46,67\% .2 = 6,66\% \)

\(\overline {{M_A}}  = \frac{{{m_A}}}{{{n_A}}} = \frac{{100}}{{2,6786}} = 37,33 \to {d_{A/{H_2}}} = \frac{{37,33}}{2} = 18,665\)