HOC24
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C2H4 + Br2==>C2H4Br2
metan(CH4) ko td với brom
b, mol hỗn hợp = 0.15(mol)
Do chỉ có etilen td được với brom nên khí thoát ra là metan
==>mol Ch4=0.1(mol)==>molC2H4=0.05(mol)
%V CH4=66.67%==>%VC2H4=33.33(bài này dễ mà )
kết tủa trắng là 2,4,6 - tribromphenol==> mol kết tủa=19,86:331=0,06(mol)
C2H5OH + Na==> C2H5ONa + 1/2H2
C6H5OH + Na ==> C6H5ONa + 1/2H2
C6H5OH + 3Br2 ==> C6H2OHBr3(kết tủa trắng) + 3HBr
Gọi mol C2H5OH,C6H5OH lần lượt là a,b
Có mol H2=0,15(mol)=1/2a+1/2b
b=0.06(mol)
a=0,24(mol),b=0,06(mol)
==>molC2H5OH=0,24(mol),mol phenol=0,06(mol)
%m C2H5OH=66,18%,%m C6H5OH=33.82%
câu 1: N Ag=0,3(mol)
NAg:Nx=0,3:0,1=3(mol)>2==>hỗn hợp phải gồm anđehit fomic ct:HCHO và 1 anđehit no X kế tiếp ==>ctpt: C2H4O
Đề bài có cho kế tiếp không?
ta tính được nco2=0,4(mol)
mol h2o=0,5(mol)
từ 2 điều trên , ta suy ra đây là ancol no , đơn chức , mạch hở nên mol ancol=0,5-0.4=0,1
ctct:CnH2n+1OH
n=mol CO2:mol ancol=0,4:0,1=4. Suy ra ctpt:C4H9OH
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