Đáp án:

 A là C2H6

V kk=49 lít

Giải thích các bước giải:

\[\begin{gathered}
  A + {O_2}\xrightarrow{{}}C{O_2} + {H_2}O \hfill \\
  {n_A} = \frac{{0,672}}{{22,4}} = 0,03{\text{ mol}} \to {\text{M}} = \frac{{0,9}}{{0,03}} = 30 \hfill \\
  {n_{C{O_2}}} = \frac{{11}}{{44}} = 0,25{\text{ mol; }}{{\text{n}}_{{H_2}O}} = \frac{{6,75}}{{18}} = 0,375{\text{ mol}} \hfill \\
  {{\text{n}}_C} = {n_{C{O_2}}} = 0,25{\text{ mol; }}{{\text{n}}_H} = 2{n_{{H_2}O}} = 0,75{\text{ mol}} \hfill \\
   \to {{\text{n}}_C}:{n_H} = 0,25:0,75 = 1:3 \hfill \\
   \to A:{(C{H_3})_n} \to 15n = 30 \to n = 2 \to A:{C_2}{H_6} \hfill \\
  2{C_2}{H_6} + 7{O_2}\xrightarrow{{}}4C{O_2} + 6{H_2}O \hfill \\
  BT\;O:2{n_{{O_2}}} = 2{n_{C{O_2}}} + {n_{{H_2}O}} \to {n_{{O_2}}} = 0,4375{\text{ mol}} \to {{\text{V}}_{{O_2}}} = 0,4375.22,4 = 9,8{\text{ lit}} \hfill \\
   \to {V_{kk}} = 9,8.5 = 49{\text{ lit}} \hfill \\ 
\end{gathered} \]

a,

$n_C=n_{CO_2}=\dfrac{11}{44}=0,25 mol$

$n_H=2n_{H_2O}=\dfrac{2.6,75}{18}=0,75 mol$

$n_C : n_H= 0,25:0,75=1:3$

$\Rightarrow$ CTĐGN $(CH_3)_n$

$n_X=\dfrac{0,672}{22,4}=0,03 mol$

$\Rightarrow M_X=\dfrac{0,9}{0,03}=30$

$\Rightarrow n=2$

Vậy CTPT X là $C_2H_6$

b,

$2C_2H_6+7O_2\buildrel{{t^o}}\over\to 4CO_2+6H_2O$

c,

Theo PTHH, $n_{O_2}=\dfrac{7}{4}n_{CO_2}=0,4375 mol$

$\Rightarrow V_{kk}=5V_{O_2}=5.0,4375.22,4=49l$