Đáp án:

\(A\) nặng gấp 2 lần \(B\)

Giải thích các bước giải:

 Ta có:

\({m_A} = {m_{{O_2}}} + {m_{C{O_2}}} + {m_{{H_2}}}\)

\( = 1,25.32 + 0,125.44 + 2,5.2 = 50,5{\text{ gam}}\)

\( \to {n_{hh{\text{ A}}}} = 1,25 + 0,125 + 2,5 = 3,875{\text{ mol}}\)

\( \to \overline {{M_A}}  = \frac{{50,5}}{{3,875}} = 13{\text{ đvC}}\)

\({m_B} = {m_{{O_2}}} + {m_{C{O_2}}} + {m_{{H_2}O}}\)

\( = 0,25.32 + 0,125.44 + 2,5.2 = 18,5{\text{ gam}}\)

\({n_{hh{\text{B}}}} = 0,25 + 0,125 + 2,5 = 2,875{\text{ mol}}\)

\(\overline {{M_B}}  = \frac{{18,5}}{{2,875}} = 6,4348{\text{ đvC}}\)

\( \to {d_{A/B}} = \frac{{{M_A}}}{{{M_B}}} = \frac{{13}}{{6,4348}} = 2,02\)

Vậy \(A\) nặng gấp 2 lần \(B\)

$M_{A}$ = $\frac{M_{O2}.n_{O2} + M_{CO2}.n_{CO2} + M_{H2}.n_{H2}}{n_{O2} + n_{CO2} + n_{H2}}$ =$\frac{1,25.32 + 0,125.44 + 2,5.2}{1,25 + 0,125 + 2,5}$ = 13,03g/mol

$M_{B}$ = $\frac{M_{O2}.n_{O2} + M_{CO2}.n_{CO2} + M_{H2}.n_{H2}}{n_{O2} + n_{CO2} + n_{H2}}$ =$\frac{0,25.32 + 0,125.44 + 2,5.2}{0,25 + 0,125 + 2,5}$ = 6,43g/mol

→ Hỗn hợp khí B nặng hơn