\[\begin{array}{l}
nC{O_2} = \frac{{6,72}}{{22,4}} = 0,3mol\\
PTHH:{C_6}{H_{12}}{O_6} - > 2C{O_2} + 2{C_2}{H_5}OH\\
n{C_6}{H_{12}}{O_6} = \frac{{nC{O_2}}}{2} = 0,15mol\\
\to m{C_6}{H_{12}}{O_6} = 0,15.180 = 27g\\
nC{O_2} = n{C_2}{H_5}OH = 0,3mol\\
a = m{C_2}{H_5}OH = 0,3.46 = 13,8g\\
\end{array}\]