$CH_4+2O_2\overset{t^o}\to CO_2+2H_2O$
$2C_2H_2+5O_2\overset{t^o}\to 4CO_2+2H_2O$
Gọi $V_{CH_4}=a;V_{C_2H_2}=b$
Ta có :
$V_{hh}=a+b=22,4$
$V_{CO_2}=a+2b=35,84$
Ta có hpt :
$\left\{\begin{matrix}
a+b=22,4 & \\
a+2b=35,84&
\end{matrix}\right.$
$⇔\left\{\begin{matrix}
a=8,96 & \\
b=13,44 &
\end{matrix}\right.$
$a/\%V_{CH_4}=\dfrac{8,96.100\%}{22,4}=40\%$
$\%V_{C_2H_2}=100\%-40\%=60\%$
$b/V_{O_2}=2a+2,5b=51,52l$
$c/CO_2+NaOH\to Na_2CO_3+H_2O$
Theo pt :
$n_{Na_2CO_3}=n_{CO_2}=\dfrac{35,84}{22,4}=1,6mol$
$⇒C_M Na_2CO_3=\dfrac{1,6}{0,5}=3,2M$