a,
$n_{CO_2}=\dfrac{1,92}{22,4}=0,08(mol)$
$Na_2CO_3+H_2SO_4\to Na_2SO_4+CO_2+H_2O$
$\Rightarrow n_{Na_2CO_3}=0,08(mol)$
$RCOONa+NaOH\buildrel{{t^o}}\over\to Y+Na_2CO_3$
Ta có: $n_{NaOH}=n_Y=n_{Na_2CO_3}=0,08(mol)$
$\overline{M}_Y=11,5.2=23$
$\Rightarrow m_Y=0,08.23=1,84g$
BTKL:
$m=1,84+0,08.106-0,08.40=7,12g$
b,
$\overline{M}_Y=23$
$\Rightarrow$ có ankan có $M<23$
$\Rightarrow$ có $CH_4$
Vậy trong 3 ankan, có 1 ankan tên là metan.