Cho `0,45(mol)` $\begin{cases}C_2H_2:x(mol)\\ H_2:y(mol)\\\end{cases}$
Ta có: `x+y=0,45(mol)(1)`
`C_2H_2+H_2\overset{xt, t^o}{\to}C_2H_4`
`C_2H_2+2H_2\overset{t^o, xt}{\to}C_2H_6`
Cho $\begin{cases} C_2H_4:a(mol)\\ C_2H_6: b(mol)\\\end{cases}$
Ta có: `a+2b=y(mol)`
`n_{C_2H_2\ dư}=x-(a+b) (mol)`
Mặc khác:
`∑_{Y}=4.7,125`
`=>28,5=\frac{m_{C_2H_2\ dư}+m_{C_2H_4}+m_{C_2H_6}}{m_{C_2H_2\ dư}+m_{C_2H_4}+m_{C_2H_6}}`
`<=>28,5=\frac{26(x-a-b)+28a+30b}{x-a-b+a+b}`
`<=>28,5=\frac{26x-26a-26b+28a+30b}{x}`
`<=>26x+2a+4b=28,5x`
`<=>2(a+2b)=2,5x`
`<=>2,5x-2y=0(2)`
Từ `(1),(2)=>`$ \begin{cases}x=0,2(mol)\\ y=0,25(mol)\\\end{cases}$
`=> %V_{C_2H_2}=\frac{0,2.100%}{0,45}\approx 44,4%`
`%V_{H_2}=\frac{0,25.100%}{0,45}\approx 55,55%`
