\(n_C=n_{CO_2}=\frac{13,2}{44}=0,3\left(mol\right)\)
\(n_H=2n_{H_2O}=\frac{2.1}{18}=\frac{1}{9}\left(mol\right)\)
\(\Rightarrow m_O=4,5-(0,3.12+\frac{1}{9}.1)=\frac{71}{90}\left(g\right)\)
=> CT: CxHyOz
Ta có x:y:z=0,3:\(\frac{1}{9}:\frac{71}{90}\)=2,7 : 1 : 7,1
MX = 30
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