Gọi \(\left\{{}\begin{matrix}n_A=x\left(mol\right)\\n_B=y\left(mol\right)\end{matrix}\right.\)
`=>` \(x.\left(14n+16\right)+y.\left(14n+18\right)=13,2\)
`=> 14xn + 16x + 14yn + 18y = 13,2`
`=> 14(xn + yn) + 16x + 18y = 13,2 (1)`
Theo BTNT C: \(n_{CO_2}=n.n_A+n.n_B=xn+yn=\dfrac{29,7}{44}=0,675\left(mol\right)\)
`=> xn + yn = 0,675(2)`
Thay `(2)` vào `(1)`
`=> 14.0,675 + 16x + 18y = 13,2`
`=> 16x + 18y = 3,75`
`=>` \(\dfrac{3,75}{18}< x+y< \dfrac{3,75}{16}\)
`=>` \(0,208n< xn+yn< 0,23n\)
`=>` \(0,208n< 0,675< 0,23n\)
`=>` \(3,245>n>2,9\)
`=> n = 3`
`=> A, B` lần lượt là \(C_3H_6O,C_3H_8O\)
`- C_3H_6O:`
`+ CH_3 - CH_2 - CHO`
`+ CH_3 - CO - CH_3`
`- C_3H_8O:`
`+ CH_3 - CH_2 - CH_2 - OH`
`+ CH_3 - CH(OH) - CH_3`
`+ C_2H_5 - O - CH_3`