$n_{H_2}:n_{C_2H_4}=3a:2a$

`->`$n_X=5a(mol)$

$m_X=2 \times 3a+28 \times 2a=62a(gam)$

`->`$M_X=\dfrac{m_X}{n_X}=\dfrac{62a}{5a}=\dfrac{62}{5}(g/mol)$

$m_X=m_Y=62a(gam)$

`->`$\dfrac{M_X}{M_Y}=\dfrac{\dfrac{62}{5}}{\dfrac{62a}{n_Y}}=0.8$

`<=>`$\dfrac{n_Y}{5a}=0.8$

`=>`$n_Y=4a(mol)$

$n_X-n_Y=n_{C_2H_4,pu}=5a-4a=a(mol)$

$n_{H_2}>n_{C_2H_4}$`->` Tính theo $C_2H_4.$

$\%H=\dfrac{n_{C_2H_4,pu}}{n_{C_2H_4}} \times 100\%=\dfrac{a}{2a} \times 100\%=50\%$

Cho `n_{H_2\ bd}=3(mol);n_{C_2H_4\ bd}=2(mol)`

`->M_X={3.2+2.28}/{3+2}=12,4(g//mol)`

`->M_Y={12,4}/{0,8}=15,5(g//mol)`

Đặt `n_{C_2H_6}=x(mol)`

`C_2H_4+H_2` $\xrightarrow{t^o,Ni}$ `C_2H_6`

BĐ: `2`   `3(mol)`

PU: `x←x←x(mol)`

CB: `2-x`    `3-x`    `x(mol)`

`->M_Y={28(2-x)+2(3-x)+30x}/{2-x+3-x+x}=15,5`

`->x=1(mol)`

`2<3->H` theo `C_2H_4`

`H=1/{2}.100\%=50\%`