Answer:
About 0.164 g of glucose.
Explanation:
We can determine the mass of glucose produced given the volume of oxygen gas produced with stoichiometry.
Recall that at STP, a mole of any gas occupies a volume of 22.4 L.
From the reaction, six moles of oxygen gas is produced for every one mole of glucose.
Lastly, the molecular weight of glucose is 180.18 g/mol.
Therefore:
[tex]\displaystyle \begin{aligned} 122\text{ mL O$_2$} & \cdot \frac{1\text{ L O$_2$}}{1000\text{ mL O$_2$}}\cdot \frac{1\text{ mol O$_2$}}{22.4\text{ L O$_2$}} \cdot \frac{1\text{ mol C$_6$H$_{12}$O$_6$}}{6\text{ mol O$_2$}} \cdot \frac{180.18\text{ g C$_6$H$_{12}$O$_6$}}{1\text{ mol C$_6$H$_{12}$O$_6$}} \\ \\ & = 0.164\text{ g C$_6$H$_{12}$O$_6$}} \end{aligned}[/tex]
Therefore, about 0.164 g of glucose is produced.
