a panmitic: `C_(15)H_(31)COOH`
a stearic: `C_(17)H_(35)COOH`
a oleic: `C_(17)H_(33)COOH`
a linoleic: `C_(17)H_(31)COOH`
Béo `+ 3NaOH -> 3` muối Béo `+` Glixerol
`1. A`
`m` Béo `= 2,225.80%=1,78` (kg)
`=> n` Béo `= (1,78)/(890)=0,002` (mol)
`nGli = n` Béo `= 0,002` (mol)
`=> m Gli = 0,184` (kg)
`2. B`
`n Béo = x` (mol)
* Vì có 2 muối `=>` loại `A, D`
* Tr/h 1: có 2 gốc `C_(15)H_(31)COO`
`=> nC_(15)H_(31)COONa = 2x`
`=> m C_(15)H_(31)COONa= 278.2x=556x`
`=> nC_(17)H_(35)COONa = x`
` => m C_(17)H_(35)COONa = 306x`
`=>` Tỉ lệ muối `: (556x)/(306x)=1,8167` ( t/m)
* Tr/h 2: có 2 gốc `C_(17)H_(35)COO`
`=> nC_(17)H_(35)COONa = 2x`
`=> m C_(17)H_(35)COONa= 306.2x=612x`
` => nC_(15)H_(31)COONa = x`
`=> m C_(15)H_(31)COONa = 278x`
`=>` Tỉ lệ muối `: (612x)/(278x)=2,201` ( Loại)
`3. D`
` nGli = (46)/(92)=0,5` (mol)
` nGli = n` Béo `= 0,5` (mol)
`=> M` Béo `= 888`
Béo gồm `: 2` gốc `R_1COO; 1` gốc `R_2COO`
`=> 2(R_1+44) + R_2+44 + C_3H_5 = 888`
` => 2R1 + R2 = 715`
Ta thấy : `R_1 = C_(17)H_(35); R_2= C_(17)H_(33)`
