\(n_{C_2H_5}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
\(\rightarrow m_{C_2H_5}=0,05.29=1,45\left(g\right)\)
\(\rightarrow\%m_{C_2H_5}=\frac{1,45}{7,7}.100\%\approx18,83\%\)
\(m_{C_2H_4,C_3H_4}=7,7-1,45=6,25\left(g\right)\)
gọi số mol của etilen là a;propin là b ta có:
26a+40b=6,25(1)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_3H_4+2Br_2\rightarrow C_3H_4Br_4\)
\(n_{Br_2}=\frac{40}{160}=0,25\left(mol\right)=n_{C_2H_4}+2n_{C_3H_4}\)
\(\Leftrightarrow a+2b=0,25\left(2\right)\)
từ (1) và (2) ta có hệ:
\(\left\{{}\begin{matrix}26a+40b=6,25\\a+2b=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\frac{5}{24}\\b=\frac{1}{48}\end{matrix}\right.\)
\(\%m_{C_2H_4}=\frac{\frac{5}{24}.26}{7,7}.100\%\approx70,35\%\)
\(\%m_{C_3H_4}=\frac{\frac{1}{48}.40}{7,7}.100\%\approx10,82\%\)
Khi thoát ra ra là Etan
\(n_{C_2H_6}= \dfrac{1,12}{22,4} = 0,05(mol)\)
Gọi \(n_{C_2H_4} =a (mol) ; n_{C_3H_4} = b(mol) \Rightarrow 28a +40 b = 7,7-0,05.30 = 6,2(1)\)
\(C_2H_4 + Br_2 \to C_2H_4Br_2\\ C_3H_4 + 2Br_2 \to C_3H_4Br_4\\ n_{Br_2} = a + 2b = \dfrac{40}{160}=0,25(2)\\ (1)(2)\Rightarrow a = 0,15 ; b = 0,05\\ \%V_{C_2H_6} = \%V_{C_3H_4} = \dfrac{0,05}{0,05 + 0,15 + 0,05}.100\% = 20\%\\ \%V_{C_2H_4} = \dfrac{0,15}{0,05 + 0,05 + 0,15}.100\% = 60\%\)
