Câu 12. C6H5NH2 + 3Br2 = C6H2Br3NH2 + 3HBr
-----------------------------1/25<-----1/75mol
n(2,4,6...) = 4,4/330 = 1/75
=> mBr2 = 1/25x 160 = 6,4g
=> mddBr2 = 6,4:3x100=640/3 gam
=> VddBr2 = m/D= 640/3 : 1,3 =164,1ml
Câu 13. (CH3)3-NH-NO3 + NaOH = (CH3)3-N + NaNO3 + H2O
nNaNO3= 13,6/85=0,16mol = n (CH3)3-N
=> m(CH3)3N = 0,16x59=9,44g
Câu 14. m HCl = m muối - m amin= 12,415 - 7,67= 4,754g
=> n HCl=0,13mol = n amin
=> Mamin = 7,67/0,13= 59 ( C3H9N) => 4 đồng phân