\(n_{CaCO_3}=n_{CO_2}=0,5mol\\ \Delta m_{dd}=-15,4=0,5.44+18n_{H_2O}-50\\ n_{H_2O}=0,7mol\\ n_{H_2O}>n_{CO_2},đốt.C_2H_2.có:n_{H_2O}< n_{CO_2}\\ \Rightarrow Z:ankan\left(C_nH_{2n+2}\right)\\ n_A=\dfrac{8,96}{22,4}=0,4mol\\ n_{trung.bình.A}=\dfrac{0,5}{0,4}=1,25\\ \Rightarrow n=1\\ Z:CH_4\)
\(n_{CH_4}=a\\ n_{C_2H_2}=b\\ a+b=0,4\\a+2b=0,5\\ a=0,3;b=0,1\\ \Rightarrow6,72L.A.có:n_{C_2H_2}=0,1\cdot\dfrac{6,72}{8,96}=0,075mol\\C_2H_2+2Br_2->C_2H_2Br_4\\ n_{Br_2}=\dfrac{20}{160}=0,125mol\\ Có:\dfrac{n_{Br_2}}{2}< n_{C_2H_2}\\ \Rightarrow Sp:C_2H_{2\left(dư\right)}:0,075-0,125:2=0,0125mol\\ m_{C_2H_2}=0,0125\cdot26=0,325g\\ m_{C_2H_2Br_4}=0,0625\cdot186=11,625g\)
