hh gồm axetilen và metan nhưng chỉ có axetilen pứ:
\(n_{Br_2}=\frac{32}{160}=0,2\left(mol\right)\)
\(PTHH:C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
(mol)____0,1_______0,2_______0,1__
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\frac{0,1.22,4}{6,72}.100\%=33,3\left(\%\right)\\\%V_{CH_4}=100-33,3=66,7\left(\%\right)\end{matrix}\right.\)
\(n_{CH_4}=\frac{6,72-2,24}{22,4}=0,2\left(mol\right)\)
\(PTHH:CH_4+Cl_2\underrightarrow{as}CH_3Cl+HCl\)
(mol)____0,2___0,2_________________
\(V_{Cl_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{hh}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Ta có:
\(n_{Br2\left(pư\right)}=\frac{64}{160}=0,4\left(mol\right)\Rightarrow n_{C2H2}=\frac{1}{2}n_{Br2}=0,2\left(mol\right)\)
\(\Rightarrow n_{CH4}=0,3-0,2=0,1\left(mol\right)\)
Vì % số mol=% thể tích
\(\%V_{CH4}=\frac{0,1}{0,3}=33,33\%\Rightarrow\%V_{C2H2}=66,67\%\)
\(CH_4+Cl_2\rightarrow CH_3Cl+HCl\)
\(\Rightarrow n_{Cl2}=n_{CH4}=0,1\left(mol\right)\Rightarrow V_{Cl2}=0,1.22,4=2,24\left(l\right)\)
Không cày nữa đi chơi game :)Hùng Nguyễn
