\(a.PTK_A=29.2=58\left(đ.v.C\right)\\ Đặt:C_xH_y\left(x,y:nguyên,dương\right)\\ y=\dfrac{17,24\%.58}{1}=10\\ x=\dfrac{58-10.1}{12}=4\\ \Rightarrow CTHH:C_4H_{10}\\ b.2C_4H_{10}+13O_2\underrightarrow{^{to}}8CO_2+10H_2O\\ n_{CO_2}=\dfrac{26,4}{44}=0,6\left(mol\right)\\ \Rightarrow n_{C_4H_{10}}=\dfrac{2}{8}.0,6=0,15\left(mol\right)\\ V_{A\left(đktc\right)}=V_{C_4H_{10}\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)
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