Đáp án:
a) $\%m_{{C_2}{H_5}COOH}=74\%;\ \%m_{C{H_3}CHO}=13,2\%;\ \%m_{C{H_3}OH}=12,8\%$
b) $V_{CO_2}=2,24\ l$
Giải thích các bước giải:
Hỗn hợp X gồm: \(\left\{ \begin{array}{l}
{C_2}{H_5}COOH:a\\
C{H_3}OH:b\\
C{H_3}CHO:c
\end{array} \right.\)
Ta có: \({n_{Ag}} = 0,12\ mol,{n_{KOH}} = 0,2\ mol\)
Khi cho X + KOH: Chỉ có axit propionic phản ứng:
\({C_2}{H_5}COOH + KOH \to {C_2}{H_5}COOK + {H_2}O\)
\(n_{{C_2}{H_5}COOH}=n_{KOH} =a= 0,2\ mol→m_{{C_2}{H_5}COOH}=14,8\ g\)
Khi cho X + $AgNO_3/NH_3$: Chỉ có andehit axetic tham gia phản ứng:
\(C{H_3}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to C{H_3}{\rm{COON}}{{\rm{H}}_{\rm{4}}}{\rm{ + 2Ag}} \downarrow {\rm{ + 2N}}{{\rm{H}}_{\rm{4}}}{\rm{N}}{{\rm{O}}_{\rm{3}}}\)
Theo PTHH: \(n_{C{H_3}CHO}=\dfrac{1}{2}n_{Ag}=0,06\ mol→m_{C{H_3}CHO}=2,64\ g\)
a/
$\%m_{{C_2}{H_5}COOH}=\dfrac{14,8}{20}.100\%=74\%$
$\%m_{C{H_3}CHO}=\dfrac{2,64}{20}.100\%=13,2\%$
$\%m_{C{H_3}OH}=100-74-13,2=12,8\%$
b/
X + $CaCO_3$
PTHH: \(2{C_2}{H_5}COOH + CaC{O_3} \to {({C_2}{H_5}COO)_2}Ca + {H_2}O + C{O_2} \uparrow \)
Ta có: $n_{CO_2}=\dfrac{1}{2}n_{{C_2}{H_5}COOH}=0,1\ mol$
$→V_{CO_2}=0,1.22,4=2,24\ l$
