Đáp án:

Giải thích các bước giải:

$a)$

Giả sử có 1 (mol) hỗn hợp khí $A$

$M_A=7,5.2=15(g/mol)$

$\dfrac{n_{C_2H_4}}{n_{H_2}}=\dfrac{15-2}{28-15}=\dfrac{1}{1}$

Vậy $n_{C_2H_4}=n_{H_2}=0,5(mol)$

$\to \%V_{C_2H_4}=\%V_{H_2}=50\%$

`C_2H_4 + H_2`$\xrightarrow[t°]{Ni}$`C_2H_6`

Giả sử có `a` `(mol)` `C_2H_6` tạo thành 

`→ B` gồm: 

$\begin{cases} C_2H_6: a(mol)\\C_2H_4: 0,5 - a (mol)\\H_2: 0,5 - a(mol)\end{cases}$

Ta có: 

`M_B = {30a + (0,5-a). 2 + (0,5-a) . 28}/{a+0,5-a+0,5-a}`

`= 9 . 2` 

`→ a = 1/6 (mol)`

`→` $\begin{cases} C_2H_6: 1/6(mol)\\H_2: 1/3(mol)\\C_2H_4: 1/3(mol) \end{cases}$

`%V_{H_2} = 1/3 : 5/6 . 100 = 40%` 

`%V_{C_2H_6} = 1/6 : 5/6 . 100 = 20%` 

`%V_{C_2H_4} = 100 - 20 - 40 = 40%`

$b)$

$\dfrac{M_A}{M_B}=\dfrac{n_B}{n_A}$

$\to n_{B}=\dfrac{M_A.n_A}{M_B}=\dfrac{7,5.1.2}{9.2}=\dfrac{5}{6}(mol)$

$H=\dfrac{1-\dfrac{5}{6}}{0,5}=33,33\%$

$c)$

$C_2H_4+3O_2\xrightarrow{t^o}2CO_2+2H_2O$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

$ 2C_2H_6 + 7O_2\xrightarrow{t^o} 4CO_2 + 6H_2O$

`V_{C_2H_4} = 40% .  26,88 = 10,752(l)` 

`n_{C_2H_4} = {10,752}/{22,4}= 0,48(mol)` 

`%V_{H_2} = %V_{C_2H_4} `

`→ n_{H_2} = n_{C_2H_4} = 0,48(mol)` 

`V_{C_2H_6} = 20% . 26,88 = 5,376(l)` 

`n_{C_2H_6} = {5,376}/{22,4} = 0,24(mol)` 

`n_{O_2} = 1/2. 0,48 + 3 . 0,48 + 7/2 . 0,24`

`= 2,52(mol)` 

`V_{O_2} = 2,52 . 22,4 =56,448l`

Đáp án:

 

Giải thích các bước giải:

Đặt $n_{H_2}=x(mol);n_{C_2H_4}=y(mol)$

$\to \dfrac{2x+28y}{x+y}=7,5.2$

$\to x=y$

Giả sử $x=y=1(mol)$

$\to n_{H_2}=n_{C_2H_4}=1(mol)$

$\to n_A=2(mol)$

$PTHH:C_2H_4+H_2\xrightarrow[Ni]{t^o}C_2H_6$

Giả sử $n_{C_2H_6}=a(mol)$

$\to B$ gồm $\begin{cases} C_2H_6:a(mol)\\ C_2H_4:1-a(mol)\\ H_2:1-a(mol) \end{cases}$

$\to \dfrac{30a+28(1-a)+2(1-a)}{a+1-a+1-a}=9.2$

$\to \dfrac{30}{2-a}=18$

$\to a=\dfrac{1}{3}\to B:\begin{cases} C_2H_6:\dfrac{1}{3}(mol)\\ C_2H_4:\dfrac{2}{3}(mol)\\ H_2:\dfrac{2}{3}(mol) \end{cases}$

$\to n_B=\dfrac{5}{3}(mol)$

$a\bigg)$

Trong A:

$\%V_{H_2}=\%V_{C_2H_4}=\dfrac{1}{2}.100\%=50\%$

Trong B:

$\%V_{C_2H_6}=\dfrac{\dfrac{1}{5}}{\dfrac{5}{3}}.100\%=20\%$

$\%V_{C_2H_4}=\%V_{H_2}=\dfrac{2}{3}{\dfrac{5}{3}}.100\%=40\%$

$b\bigg)$

$H=\dfrac{\dfrac{1}{3}}{1}.100\%\approx 33,33\%$

$c\bigg)$

$n_B=\dfrac{26,88}{22,4}=1,2(mol)$

$\to \begin{cases} n_{C_2H_6}=0,24(mol)\\ n_{C_2H_4}=n_{H_2}=0,48(mol) \end{cases}$

$n_{O_2}=3,5n_{C_2H_6}+3n_{C_2H_4}+0,5n_{O_2}=2,52(mol)$

$\to V_{O_2}=2,52.22,4=56,448(l)$