Giải bài 10 trang 204 - SGK môn Hóa học lớp 11 nâng cao

${{m}_{{{C}_{6}}{{H}_{14}}}}=\dfrac{43\%}{100\%}.1=0,43\,gam\Rightarrow {{n}_{{{C}_{6}}{{H}_{14}}}}=0,005\,mol \\ $

${{m}_{{{C}_{7}}{{H}_{16}}}}=\dfrac{49,5\%}{100\%}.1=0,495gam\Rightarrow {{n}_{{{C}_{6}}{{H}_{14}}}}=0,00495\,mol \\ $

${{m}_{{{C}_{5}}{{H}_{12}}}}=\dfrac{1,8\%}{100\%}.1=0,018\,gam\Rightarrow {{n}_{{{C}_{5}}{{H}_{12}}}}=0,00025\,mol \\ $

${{m}_{{{C}_{8}}{{H}_{18}}}}=1-0,43-0,495-0,018=0,057\,gam\Rightarrow {{n}_{{{C}_{8}}{{H}_{18}}}}=0,0005\,mol \\ $

${{C}_{6}}{{H}_{14}}+\dfrac{19}{2}{{O}_{2}}\xrightarrow{{{t}^{0}}}6C{{O}_{2}}+7{{H}_{2}}O \\ $

${{C}_{7}}{{H}_{16}}+11{{O}_{2}}\xrightarrow{{{t}^{0}}}7C{{O}_{2}}+8{{H}_{2}}O \\ $

${{C}_{5}}{{H}_{12}}+8{{O}_{2}}\xrightarrow{{{t}^{0}}}5C{{O}_{2}}+6{{H}_{2}}O \\ $

${{C}_{8}}{{H}_{18}}+\dfrac{15}{2}{{O}_{2}}\xrightarrow{{{t}^{0}}}8C{{O}_{2}}+9{{H}_{2}}O \\ $

${{n}_{{{O}_{2}}}}=\dfrac{19}{2}{{n}_{{{C}_{6}}H{_{14}}}}+11{{n}_{{{C}_{7}}{{H}_{16}}}}+8{{n}_{{{C}_{5}}{{H}_{12}}}}+\dfrac{15}{2}{{n}_{{{C}_{8}}{{H}_{18}}}}=0,1077\,mol \\ $

$\Rightarrow {{V}_{{{O}_{2}}}}=2,41248\,L \\ \\ $

$\Rightarrow {{n}_{C{{O}_{2}}}}=6{{n}_{{{C}_{6}}H{_{14}}}}+7{{n}_{{{C}_{7}}{{H}_{16}}}}+5{{n}_{{{C}_{5}}{{H}_{12}}}}+8{{n}_{{{C}_{8}}{{H}_{18}}}}=0,0699\,mol \\ $

$\Rightarrow {{V}_{C{{O}_{2}}}}=1,56576\,L \\ $

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