a, CnH2n+2O +(3n/2)O2 -> nCO2 + (n+1)H2O

                                                0.06          0.08         (mol)

C/H = 0.06/(0.08*2) =3/8

=> CTPT: C3H8O

CTCT: CH3CH2CH2OH(propan-1-ol) ; CH3CH(OH)CH3(propan-2-ol)

b, C3H8O + 9/2 O2 -> 3CO2 + 4H2O

     0.02    <-   0.09     <- 0.06       0.08    (mol)

VO2= 0.09*22.4 =2.016(l)

m=0.02*(12*3+8+16) = 1.2(g)

c, C3H8O + CuO -> andehit => ancol bậc 1 => X: CH3CH2CH2OH

CH3CH2CH2OH + CuO ->CH3CH2CHO + H20 + Cu

   0.02                              ->   0.02 (mol)

m andehit= 0.02*(12*3+6+16) = 1.16(g)

Đáp án:

 a) $C_3H_8O$

b) 2,016l

c) 1,16g

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
{n_{C{O_2}}} = \dfrac{{2,64}}{{44}} = 0,06mol\\
{n_{{H_2}O}} = \dfrac{{1,44}}{{18}} = 0,08\,mol\\
{n_X} = {n_{{H_2}O}} - {n_{C{O_2}}} = 0,08 - 0,06 = 0,02\,mol\\
C = \dfrac{{{n_{C{O_2}}}}}{{{n_X}}} = \dfrac{{0,06}}{{0,02}} = 3\\
CTPT:{C_3}{H_8}O\\
CTCT:\\
C{H_3} - C{H_2} - C{H_2} - OH:propan - 1 - ol\\
C{H_3} - CH(OH) - C{H_3}:propan - 2 - ol\\
b)\\
2{C_3}{H_8}O + 9{O_2} \xrightarrow{t^0} 6C{O_2} + 8{H_2}O\\
{n_{{O_2}}} = 0,02 \times \dfrac{9}{2} = 0,09\,mol\\
{V_{{O_2}}} = 0,09 \times 22,4 = 2,016l\\
c)\\
{C_3}{H_7}OH + CuO \xrightarrow{t^0} {C_2}{H_5}CHO + Cu + {H_2}O\\
{n_{{C_2}{H_5}CHO}} = {n_{{C_3}{H_7}OH}} = 0,02\,mol\\
{m_{{C_2}{H_5}CHO}} = 0,02 \times 58 = 1,16g
\end{array}\)