Step 1 - Understanding the stoichiometry of the reaction
The given reaction is:
[tex]N_{2(g)}+2O_{2(g)}\rightleftarrows N_2O_{4(g)}[/tex]Note that 1 mole of Nitrogen gas react with 2 moles of Oxygen gas thus producing 1 mole of N2O4 gas. Since this is a fixed relation, we can state that:
each 2 moles of O2 produce 1 mole of N2O4
Step 2 - Converting the relation in moles to a relation in grams
We can convert the relation in moles between O2 and N2O4 to a relation in grams by multiplying each number of moles by the respective molar masses of the substances (32 g/mol for O2; 92 g/mol for N2O4):
[tex]\begin{gathered} O_2\to2\times32=64\text{ g} \\ N_2O_4\to1\times92=92\text{ g} \end{gathered}[/tex]We can state therefore that:
64 g of O2 (2 moles) produce 92 g of N2O4
This is like a cake recipe. It is a fixed proportion in grams, and we'll be using it to solve the exercise.
Step 3 - Finding the required moles of O2
Now that we have found a "recipe" for the reaction, we can use it to predict how much O2 would be needed. Let's remember that, in step 2, we have discovered that 2 moles of O2 produce 92 g of N2O4.
Therefore, to produce 23.6 grams, we can set the following proportion:
[tex]\begin{gathered} 2\text{ moles of O2 produce ---- 92 g of N2O4} \\ x\text{ ------------------ 23.6 g of N2O4} \\ \\ x=\frac{23.6\times2}{92}=0.51\text{ moles of O2} \end{gathered}[/tex]Therefore, 0.51 moles of O2 would be required.
