\(a,n_A=\dfrac{1,344}{22,4}=0,06mol\\ m_A=0,06.72=4,32g\\ m_H=2\cdot\dfrac{6,48}{18}=0,72g\\ m_C=4,32-0,72=3,6g\)
Gọi CTPT HC A \(C_xH_y\)
Ta có:
\(\dfrac{12x}{3,6}=\dfrac{y}{0,72}=\dfrac{72}{4,32}\\ \Rightarrow x=5;y=12\)
Vậy CTPT HC A \(C_5H_{12}\)
\(b,C_5H_6+8O_2\xrightarrow[t^0]{}5CO_2+6H_2O\\ n_{O_2}=8n_{O_2}=0,48mol\\ V_{O_2}=0,48.22,4=10,752l\)