\(BTKL:m_D+m_{O_2}=m_{CO_2}+m_{H_2O}\\ < =>m_D+14,4=13,2+7,2\\ < =>m_D=6g\\ M_D=\dfrac{6}{0,1}=60g/mol\\ b.m_C=\dfrac{13,2}{44}\cdot12=3,6g\\ m_H=\dfrac{7,2}{18}.2=0,8g\\ m_{C+H}=0,8+3,6=4,4g< m_D\\ Trong.D.có.C,H,O\\ CTPT\left(D\right):C_xH_yO_z\\ m_O=6-4,4=1,6g\\ \dfrac{12x}{3,6}=\dfrac{y}{0,8}=\dfrac{16z}{1,6}=\dfrac{60}{6}\\ =>x=3;y=8;z=1\\ =>CTPT\left(A\right)C_3H_8O\)
