\(PTHH:CO+\frac{1}{2}O_2\rightarrow CO_2\)
\(C_4H_{10}+6,5O_2\rightarrow4CO_2+5H_2O\)
Gọi số mol CO là x; CO2 là y; C4H10 là z
\(\rightarrow x+y+z=\frac{17,92}{22,4}=0,8\left(mol\right)\)
\(n_{O2}=0,5x+6,5z=\frac{75,04}{22,4}=3,35\left(mol\right)\)
\(n_{O2}=x+y+4z=\frac{51,52}{22,4}=2,3\left(mol\right)\)
\(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\\z=0,5\end{matrix}\right.\)
% Số mol=% Thể tích
\(\rightarrow\%V_{CO}=25\%;\%V_{CO2}=12,5\%;\%V_{C4H10}=62,5\%\)
Mặt khác : \(n_{H2O}=5n_{C4H10}=5z=2,5\left(mol\right)\)
\(\rightarrow m_{H2O}=a=2,5.18=45\left(g\right)\)
