Đáp án + Giải thích các bước giải:

1.

Đặt $V_{C_2H_4}=x(cm^3);V_{C_2H_2}=y(cm^3)$

$\to x+y=35(1)$

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

$2C_2H_2+5O_2\xrightarrow{t^o}4CO_2+2H_2O$

Theo PT: $3x+2,5y=V_{O_2}=90(2)$

Từ $(1)(2)\to x=5(cm^3);y=30(cm^3)$

$\%V_{C_2H_4}=\frac{5}{35}.100\%\approx 14,29\%$

$\%V_{C_2H_2}=100-14,29=85,71\%$

2.

Đặt $V_{CH_4}=x(cm^3);V_{C_2H_4}=y(cm^3)$

$\to x+y=20(1)$

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

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

Theo PT: $2x+3y=V_{O_2}=50(2)$

Từ $(1)(2)\to x=y=10(cm^3)$

$\%V_{C_2H_4}=\%V_{CH_4}=\frac{10}{20}.100\%=50\%$

Giải thích các bước giải:

$1.$

Đặt: $V_{C_2H_4}=x(cm^3),\ V_{C_2H_2}=y(cm^3)$ 

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

$C_2H_2+\dfrac{5}{2}O_2\xrightarrow{t^o}2CO_2+H_2O$

$\begin{cases} x+y=35(cm^3)\\3x+\dfrac{5}{2}y=90(cm^3) \end{cases}$

$\to x=5(cm^3),\ y=30(cm^3)$

$\%V_{C_2H_4}=\dfrac{5}{35}=14,29\%$

$\%V_{C_2H_2}=100\%-14,29\%=85,71\%$

$2.$

Đặt: $V_{CH_4}=x(cm^3),\ V_{C_2H_4}=y(cm^3)$ 

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

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

$\begin{cases} x+y=20(cm^3)\\2x+3y=50(cm^3) \end{cases}$

$\to x=y=10(cm^3)$

$\%V_{CH_4}=\dfrac{10}{20}=50\%$

$\%V_{C_2H_4}=100\%-50\%=50\%$