25 tháng 3 2024 lúc 22:23
\(n_{Br_2}=\dfrac{16,8}{160}=0,105mol\\ n_{hh}=\dfrac{1,68}{22,4}=0,075mol\\ n_{C_2H_4}=a;n_{C_2H_2}=b\\ C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ C_2H_2+2Br_2\xrightarrow[]{}C_2H_2Br_4\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,075\\a+2b=0,105\end{matrix}\right.\\ =>a=0,045;b=0,003\\ \%V_{C_2H_4}=\dfrac{0,045}{0,075}\cdot100\%=60\%\\ \%V_{C_2H_2}=40\%\)
Đúng 2
Bình luận (0)
