Đáp án:
Metan : $CH_{4}$
Etilen : $C_{2}H_{4}$
Axtilen: $C_{2}H_{2}$
$nC_{2}H_{2}=\frac{0,56}{22,4}=0,025$
$nY=\frac{1,12}{22,4}=0,05$
$nCH_{4} +nC_{2}H_{4}=0,05-0,025=0,025$ (1)
$nCaCO_{3}=\frac{8,125}{100}=0,08125$
BTNT "C"
$nCH_{4} + 2nC_{2}H_{4} +2nC_{2}H_{2}=nCaCO_{3}$
⇔$nCH_{4} +2nC_{2}H_{4}=0,08125-0,025.2=0,03125$(2)
(1)(2)⇒$\left \{ {{nCH_{4} =0,01875} \atop {nC_{2}H_{4}=0,00625}} \right.$
$V\%CH_{4}=n\%CH_{4}=\frac{0,01875}{0,05}.100=37,5\%$
$V\%C_{2}H_{2}=\frac{0,025}{0,05}=50\%$
$V\%C_{2}H_{4}=100-37,5-50=12,5\%$