4)C2H4 + Br2 ---> C2H4Br2
nBr2 = 8: 160= 0,05 mol
=> nC2H4= 0,05 mol
=> VC2H4= 0,05.22,4 = 1,12 (l)
=> % VC2H4= 1,12\3,36.100% ≈≈ 33%
=> % VCH4 = 67%
Câu 4:
Metan CH4
Etilen: C2H4
\(n_{hh}=0,15\left(mol\right)\)
\(n_{Br2}=\frac{8}{160}=0,05\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,05________0,05 _________
\(V\%_{C2H4}=n\%_{C2H4}=\frac{0,05}{0,15}.100\%=33,33\%\)
\(\Rightarrow V\%_{CH4}=100\%-33,33\%=66,67\%\)
Câu 5:
Đề thiếu dữ kiện hh ban đầu
Propan: C3H8
Propin: C3H4
\(CTCT:CH\equiv C-CH_3\)
\(CH\equiv C-CH_3+AgNO_3+NH_3\rightarrow CAg\equiv C-CH_3+NH_4NO_3\)
\(n_{C3H3Ag}=\frac{1,47}{147}=0,01\left(mol\right)\)
\(\Rightarrow n_{C3H4}=n_{C3H3Ag}=0,01\left(mol\right)\)
\(\Rightarrow m_{C3H4}=0,01.40=0,4\left(g\right)\)
