Đốt cháy hoàn toàn 14,28 gam

Chọn đáp án A

Giải thích:

$\begin{array}{l}+\left\{\begin{array}{l}{n}_C=n_{C{O}_2}=\mathrm{0,58}\\ n_H=2n_{H_{2}O}=\mathrm{0,92}\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_O=(\mathrm{14,28}-\mathrm{0,58.12}-\mathrm{0,92}):16=\mathrm{0,4}\\ n_X=\mathrm{0,5}{n}_O=\mathrm{0,2}\end{array}\right.\\ +1<\frac{n_{KOH}}{n_X}=\frac{\mathrm{0,23}}{\mathrm{0,2}}<2\Rightarrow Xgoàm\left\{\begin{array}{l}Ylaestecuaancol\left(xmol\right)\\ Zlaestecuaphenol\left(ymol\right)\end{array}\right.\Rightarrow \left\{\begin{array}{l}x+y=\mathrm{0,2}\\ x+2y=\mathrm{0,23}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,17}\\ y=\mathrm{0,03}\end{array}\right.\\ +{\overline{C}}_X=\frac{n_{C{O}_2}}{n_X}=\frac{\mathrm{0,58}}{\mathrm{0,2}}=\mathrm{2,9}\Rightarrow YlaHCOOC{H}_3\Rightarrow C_Z=\frac{\mathrm{0,58}-\mathrm{0,17.2}}{\mathrm{0,03}}=8.\\ +X\stackrel{KOH}{\to }2muoi\Rightarrow ZlaHCOO{C}_6{H}_{4}C{H}_3;haimuoila\left\{\begin{array}{l}HCOOK:\mathrm{0,2}mol\\ OK{C}_6{H}_{4}C{H}_3:\mathrm{0,03}mol\end{array}\right.\\ \Rightarrow \%HCOOK=\frac{\mathrm{0,2.84}}{\mathrm{0,2.84}+\mathrm{0,03.146}}=\mathrm{79,32}\%.\end{array}$