`#Ly`

$a)$ 

$\overline{MhhX}$ $=$ $d.M_{N_2}=0,5.28=14$ $(g/mol)$

Áp dụng sơ đồ đường chéo:

$\rm \text{$V_{H_2}$ $M_{H_2}$ $=$ $2$ } \hspace{4,8cm} \text{$12$} \\\hspace{3cm} \huge{\searrow} \hspace{2,5cm}\huge{\nearrow} \\\hspace{4,5cm}\text{$\overline{M}$ $=$ $14$ } \\ \\\hspace{3cm} \huge{\nearrow} \hspace{2,5cm} \huge{\searrow} \\\ \text{$V_{C_2H_2}$ $M_{C_2H_2}$ $=$ $26$ } \hspace{4cm} \text{$12$}$

$⇒$ $\dfrac{V_{H_2}}{V_{C_2H_2}}$ $=$ $\dfrac{1}{1}$ $⇒$ $V_{H_2}=V_{C_2H_2}$

Mà $V_{H_2}+V_{C_2H_2}=17,92$ $(l)$

$V_{H_2}$ $=$ $17,92:2=8,96$ $(l)$

$V_{C_2H_2}$ $=$ $8,96$ $(l)$

$n_{H_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{8,96}{22,4}$ $=$ $0,4$ $(mol)$

$n_{C_2H_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{8,96}{22,4}$ $=$ $0,4$ $(mol)$

$b)$ 

$V_{O_2}$ $=$ $224:100.20=44,8$ $(l)$

$n_{O_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{44,8}{22,4}$ $=$ $2$ $(mol)$

PTHH: 

$(1):$ $2H_2$ $+$ $O_2$ $\xrightarrow[]{t^o}$ $2H_2O$

$(2):$ $2C_2H_2$ $+$ $5O_2$ $\xrightarrow[]{t^o}$ $4CO_2$ $+$ $2H_2O$ 

Theo PT $(1)$, $n_{O_2}$ $=$ $\dfrac{1}{2}n_{H_2}$ $=$ $0,2$ $(mol)$

Theo PT $(2)$, $n_{O_2}$ $=$ $\dfrac{5}{2}n_{C_2H_2}$ $=$ $1$ $(mol)$

$n_{O_2(dư)}$ $=$ $2-1-0,2=0,8$ $(mol)$

$⇒$ Hỗn hợp khí $Y:$ $CO_2$, $N_2$, $O_2$ dư

$V_{O_2(dư)}$ $=$ $n.22,4=0,8.22,4=17,92$ $(l)$

$V_{N_2}$ $=$ $224-44,8=179,2$ $(l)$

Theo PT $(2)$, $n_{CO_2}$ $=$ $2n_{C_2H_2}$ $=$ $0,8$ $(mol)$

$V_{CO_2}$ $=$ $n.22,4$ $=$ $0,8.22,4=17,92$ $(l)$

$V_{hhY}$ $=$ $17,92+179,2+17,92=215,04$ $(l)$

`%V_(O_2(dư)` $=$ $\dfrac{17,92}{215,04}$`.100%` $≈$ `8,333%`

`%V_(N_2` $=$ $\dfrac{179,2}{215,04}$`.100%` $≈$ $83,33$`%`

`%V_(N_2` $=$ $100-8,33-83,33=8,34$`%`

$m_{O_2(dư)}$ $=$ $n.M=0,8.32=25,6$ $(g)$

$m_{CO_2}$ $=$ $n.M=0,8.44=35,2$ $(g)$

$n_{N_2}$ $=$ $\dfrac{V}{22,4}$ $=$ $\dfrac{179,2}{22,4}$ $=$ $8$ $(mol)$

$m_{N_2}$ $=$ $n.M=8.28=224$ $(g)$

`%m_{O_2(dư)}` $=$ $\dfrac{25,6}{25,6+35,2+224}$`.100%` $≈$ $8,98$`%`

`%m_{N_2}` $=$ $\dfrac{224}{25,6+35,2+224}$`.100%` $≈$ `78,65%`

`%m_{CO_2}` $=$ $100-8,98-78,65=12,37$`%`

Giải thích các bước giải:

`a) n_X = {17,92}/{22,4} = 0,8 mol`

`M_X = 0,5.28 = 14`$g/mol$

Gọi `n_{H_2} = x(mol);n_{C_2H2}= y(mol)`

Ta có: `2x + 26y = 0,8.14`

          `x + y = 0,8`

Theo hệ PT: `x = 0,4(mol) ; y = 0,4(mol)`

`b)``V_{O_2} = 224  . 20% = 44,8l`

`n_{O_2} = {44,8}/{22,4} = 2 mol`

`V_{N_2} = 224 - 44,8 = 179,2l`

`n_{N_2} = {179,2}/{22,4} = 8 mol`

`PTHH: 2H_2 + O_2 ->^{t^o} 2H_2O`

           `2C_2H_2 + 5O_2 ->^{t^o}  4CO_2 + 2H_2O`

`n_{O_2}(hh) = {1}/{2}n_{H_2} + {5}/{2}n_{C_2H_2}`

`n_{O_2}(hh) = {1}/{2}.0,4 + {5}/{2}.0,4 = 1,2 mol`

Vậy `n_{O_2}(dư) = 2 - 1,2 = 0,8 mol`

`n_{CO_2} = 0,4 . 4/2 = 0,8 mol`

Hỗn hợp Y gồm `N_2 ; CO_2 ; O_2(dư)`

`%V_{N_2} = {8}/{8+0,8 + 0,8}.100% = 83,33%`

`%V_{CO_2} = {0,8}/{8 + 0,8 + 0,8}.100% =8,33%`

`%V_{O_2}(dư) = 100% - 83,33% - 8,33% = 8,33%`

`%m_{N_2} = {8.28}/{8.28 + 0,8.32 + 0,8.44}.100% = 78,65%`

`%m_{CO_2} = {0,8.44}/{8.28 +0,8.32 + 0,8.44}.100% = 12,36%`

`%m_{O_2}(dư) = 100% - 78,65% - 12,36% = 8,99%`