a, \(CH\equiv CH+2Br_2\rightarrow CHBr_2-CHBr_2\)
b, \(n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{C_2H_2Br_4}=n_{C_2H_2}=0,3\left(mol\right)\)
\(\Rightarrow m_{C_2H_2Br_4}=0,3.346=103,8\left(g\right)\)
c, \(n_{Br_2}=2n_{C_2H_2}=0,6\left(mol\right)\)
\(\Rightarrow C_{M_{Br_2}}=\dfrac{0,6}{0,5}=1,2\left(M\right)\)