Gọi công thức hóa học chung của hỗn hợp X là \(C_3H_x\)
Có \(\overline{M_X}=21\cdot2=42đvC\Rightarrow12\cdot3+x\cdot1=42\)
\(\Rightarrow x=6\)
\(n_X=\dfrac{1,12}{22,4}=0,05mol\)
\(C_3H_6+\dfrac{9}{2}O_2\rightarrow3CO_2+3H_2O\)
0,05 0,15 0,15
\(m_{CO_2}=0,15\cdot44=6,6g\)
\(m_{H_2O}=0,15\cdot18=2,7g\)
\(C_3H_4+4O_2\underrightarrow{^{^{t^0}}}3CO_2+2H_2O\)
\(C_3H_6+\dfrac{9}{2}O_2\underrightarrow{^{^{t^0}}}3CO_2+3H_2O\)
\(C_3H_8+5O_2\underrightarrow{^{^{t^0}}}3CO_2+4H_2O\)
\(n_X=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(\Rightarrow n_{CO_2}=3\cdot n_X=3\cdot0.05=0.15\left(mol\right)\)
\(m_{CO_2}=0.15\cdot44=6.6\left(g\right)\)
\(m_X=21\cdot2\cdot0.05=2.1\left(g\right)\)
\(\Rightarrow m_H=m_X-m_C=2.1-0.15\cdot12=0.3\left(g\right)\)
\(n_H=0.3=0.3\left(mol\right)\)
\(\Rightarrow n_{H_2O}=0.15\left(mol\right)\)
\(m_{H_2O}=0.15\cdot18=2.7\left(g\right)\)