\(n_{CO_{2}}=0,4(mol), n_{H_{2}O}=0,6(mol)\)
\(\Rightarrow n_{C}=0,4(mol),n_{H}=1,2(mol)\)
\(m_{C} + m_{H} = 0,4 . 12 + 1,2 . 1 = 6 < m_{X} = 9,2 g\)
\(⇒\) Trong X có oxi
\(m_{O}=9,2-6=3,2(g)\)
%\(C=\frac{0,4\times 12}{9,2}\times 100=52,17\)%
%\(H=\frac{1,2}{9,2}\times 100=13,04\)%
%\(O=\frac{3,2}{9,2}\times 100=34,78\)%