\(n_{C2H4}=\frac{2,8}{22,4}=0,125\left(mol\right)\)
\(n_{H2}=\frac{11,2}{22,4}=0,05\left(mol\right)\)
Do etilen là 1 anken trong phân tử chưa 1 lkπ nên
\(n_{H2}=n\pi pư=n_{C2H4}\)
Ta có nC2H4 dư = nC2H4 ban đầu - n pứ =0,125 - 0,05 = 0,075 mol
\(PTHH:C_2H_4+H_2\rightarrow C_2H_6\)
_________0,05__0,05__________
\(m_{C2H4;C2H6\left(spư\right)}=0,05.30+0,075.28=3,6\left(g\right)\)
\(\Rightarrow n_{C2H4;C2H6\left(spu\right)}=0,05+0,075=0,125\left(mol\right)\)
\(M=\frac{3,6}{0,125}=28,8\)
\(\Rightarrow d_{X/H2}=\frac{28,8}{2}=14,4\)
