Khi cho hỗn hợp amin + Brom
\(n_X=n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\)
Khi cho hỗn hợp amin + HCl
\(n_{HCl\left(bđ\right)}=0,36\left(mol\right)\\ DoHCllấydư20\%\Rightarrow n_{HCl\left(pứ\right)}=\dfrac{0,36}{120}.100=0,3\left(mol\right)\\ n_{HCl\left(pứ\right)}=n_{metylamin}+n_X=n_{metylamin}+0,2=0,3\\ \Rightarrow n_{metylamin}=0,1\left(mol\right)\\ \Rightarrow m_X=14,5-0,1.31=11,4\left(g\right)\\ \Rightarrow M_X=\dfrac{11,4}{0,2}=57\\ \Rightarrow XcóCT:CH_2=CH-CH_2-NH_2\\ \Rightarrow ChọnD\)
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