\(n_A=\dfrac{1,904}{22,4}=0,085mol\\ n_{CO_2}=\dfrac{0,896}{22,4}=0,04mol\\ \Rightarrow n_{CH_4}=0,04mol\\ C_3H_6+Br_2->C_3H_6Br_2\\ C_3H_4+2Br_2->C_3H_4Br_4\\ n_{propen}=a;n_{propin}=b\\ a+b=0,085-0,04=0,045\left(1\right)\\ a+2b=\dfrac{9,6}{160}=0,06\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow a=0,03\\ b=0,015\\ \Rightarrow m_{propin}=40.0,015=0,6g\\ Chọn.C\)
Đúng 1
Bình luận (0)
