a)
C2H4 + Br2 → C2H4Br2
C2H2 + 2Br2 → C2H2Br4
CH≡CH +2AgNO3 + 2NH3 → AgC≡CAg↓ + 2NH4NO3
b) nC2H2 = nAg2C2 = 24,24/240 = 0,101 mol
nC2H4 = (6,72−1,68)/22,4 − 0,101 = 0,124 mol
nC3H8 = 1,68/22,4 = 0,075 mol
=>%VC2H2= %nC2H2 = 0,1010,3.100% = 33,7%
=> %VC2H4 = (0,124/0,3).100% = 41,3%; %VC3H8 =100% − (33,7%+41,3%) = 25%
mX = 0,101.26 + 0,124.28 + 0,075.44 = 9,398 g
=> %mC2H2 = 0,101.269,398.100 = 27,9%
=> %mC2H4 = 0,124.289,398.100 = 36,9%
=> %mC3H8 = 100 − (27,9+36,9) = 35,2%