Đốt cháy hoàn toàn 17,16 gam

Chọn đáp án B

Giải thích:

$\left\{\begin{array}{l}{n}_X=xmol\\ n_{H_{2}O}=ymol\end{array}\right.\Rightarrow \left\{\begin{array}{l}{m}_X=\mathrm{12.1,1}+2y+16.6x=\mathrm{17,16}\\ 2x=\mathrm{1,1}-(y+\mathrm{0,04})\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,02}mol\\ y=\mathrm{1,02}mol\end{array}\right.\Rightarrow m_{muoi}=\mathrm{17,16}+\mathrm{28.0,02}=\mathrm{17,72}gam.$