\(n_{CO2}=\frac{6,6}{44}=0,15\left(mol\right)\)
\(n_{H2O}=\frac{3,6}{18}=0,2\left(mol\right)\)
\(m_C+m_H=0,15.12+0,2.2=2,2< 3\)
\(\rightarrow\) Trong A có O
\(m_C=0,15.12=1,8\left(g\right)\)
\(m_H=0,2.2=0,4\left(g\right)\)
\(m_O=0,8\left(g\right)\)
\(\%m_C=\frac{1,8}{2}.100\%=60\%\)
\(\%m_O=\frac{0,8}{3}.100\%=26,67\%\)
\(\%m_H=13,33\%\)