Question

How many liters of oxygen, at standard conditions, are required to react with 30 grams of glucose, C6H12O6?

C6H12O6 (s) + 6 O2(g) → 6 CO2 (g) + 6 H2O (g)

Answer 1

Hey Royalbug.

Start by calculating the amount of substance [mol] of glucose you would require for this experiment:
\[\large n(\text{C}_6\text{H}_{12}\text{O}_6)=\frac{ m(\text{C}_6\text{H}_{12}\text{O}_6) }{ M(\text{C}_6\text{H}_{12}\text{O}_6) }\]
The equivalent amounts glucose and oxygen is given in a 1:6 ratio:
\[\large n(\text{C}_6\text{H}_{12}\text{O}_6)=\frac{ n(\text{O}_2) }{ 6 }~ \Rightarrow~ n(\text{O}_2)=6 ~ n(\text{C}_6\text{H}_{12}\text{O}_6)\]
Lastly we will assume that oxygen behaves as an ideal gas and approximate the volume of oxygen required to produce the \(n(\text{O}_2)\):
\[\Large p V_{O_2}=n(\text{O}_2) R T\]
Here is \(p\) the pressure, \(V_{O_2}\) the volume of oxygen, \(R\) the gas constant and \(T\) the temperature. At standard conditions (STP), the experiment is carried out at 1 bar pressure (the equivalent to 100 kilopascal), and 0\(^{\circ}\)C which is equal to 273.15 Kelvin.

Solve for the volume in the equation and you are done!