\(n_{C_2H_2}=\dfrac{0.224}{22.4}=0.01\left(mol\right)\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(0.01.........0.02........0.01\)
\(m_{C_2H_2Br_4}=0.01\cdot346=3.46\left(g\right)\)
\(V_{dd_{Br_2}}=\dfrac{0.02}{2}=0.01\left(l\right)\)
a) \(C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ \)
\(b)\\ n_{C_2H_2Br_4} = n_{C_2H_2} =\dfrac{0,224}{22,4} = 0,01(mol)\\ \Rightarrow m_{C_2H_2Br_4} = 0,01.346 = 3,46\ gam\\ c)\\ n_{Br_2} = 2n_{C_2H_2} = 0,02(mol)\\ \Rightarrow V_{dd\ brom} =\dfrac{0,02}{2} = 0,01(lít)\)
Theo gt ta có: $n_{C_2H_2}=0,01(mol)$
a, $C_2H_2+2Br_2\rightarrow C_2H_2Br_4$
b, Ta có: $m_{C_2H_2Br_4}=0,01(mol)\Rightarrow m_{C_2H_2Br_4}=1,86(g)$
c, $n_{Br_2}=0,02(mol)\Rightarrow V=0,01(l)$