Question

During photosynthesis, a green plant produces 122 mL of oxygen gas at STP. What mass of glucose

(C6H12O6) is produced during this reaction?

6CO2+ 6H2O > C6H12O6+ 602

Show all steps (Answer: 0.164g)

Answer 1

About 0.164 g of glucose.

Step-by-step 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:

`\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}`

Therefore, about 0.164 g of glucose is produced.