Đáp án:
Giải thích các bước giải:
a/ $n_{O_2}=\dfrac{6,72}{22,4}=0,3\ mol$
b/ $C_2H_6+\dfrac{7}{2}O_2\xrightarrow{t^o}2CO_2+3H_2O\\1\hspace{1,5cm}\dfrac{7}{2}\hspace{2cm}2\hspace{1,5cm}3$
$⇒n_{C_2H_6}=0,3:\dfrac{7}{2}=\dfrac{3}{35}\ mol\\n_{CO_2}=0,3.2:\dfrac{7}{2}=\dfrac{6}{35}\ mol\\n_{H_2O}=0,3.3:\dfrac{7}{2}=\dfrac{9}{35}\ mol$
c/ $C_2H_6+\dfrac{7}{2}O_2\xrightarrow{t^o}2CO_2+3H_2O$
d/ $n_{C_2H_6}=0,3:\dfrac{7}{2}⇒m_{C_2H_6}=0,3:\dfrac{7}{2}.30=\dfrac{18}{7}g$
CTCT: $CH_3-CH_3$
Tính chất:
Phản ứng thế Haloen : $C_2H_6+Cl_2\xrightarrow{askt}C_2H_5Cl+HCl$
Phản ứng tách $H_2$: $C_2H_6\xrightarrow{t^o,xt}C_2H_4+H_2$
Phản ứng cháy: $C_2H_6+\dfrac{7}{2}O_2\xrightarrow{t^o}2CO_2+3H_2O$