Giả sử số mol CH4 ban đầu là 1 (mol)
=> mX = 1.16 = 16 (g)
PTHH: \(2CH_4\underrightarrow{t^o}C_2H_2+3H_2\)
x--->0,5x-->1,5x \(\left(0< x< 1\right)\)
=> nX = 1 - x + 0,5x + 1,5x = 1 + x (mol)
\(\overline{M}_X=\dfrac{16}{1+x}\left(g/mol\right)\Rightarrow8< \overline{M}_X< 16\)
=> \(2< d_{X/H_2}< 4\)
=> C