18.5 grams of Nitrogen gas and 26.7 grams of Oxygen gas form how many grams of Dinitrogen Pentoxide if the reaction is only 80% efficient? ***THE PERCENTAGE DOES NOT CONTRIBUTE TO THE SIG FIGS... ONLY THE MEASURED MASSES OF THE GASES***

29.38 grams

Explanation:

The reaction is:

2N₂(g) + 5O₂(g) → 2N₂O₅    (1)

We need to calculate the number of moles of N₂ and O₂:

Now, we need to find the limiting reactant. From reaction (1) we have that 2 moles of N₂ react with 5 moles of O₂:

If we have 0.66 moles of N₂ and we need 0.34 moles to react with O₂, then the limiting is O₂.

We can calculate the number of moles of N₂O₅ produced:

Now, we can calculate the theoretical mass of N₂O₅:

Finally, if the reaction is only 80% efficient then the mass of N₂O₅  produced is:

Therefore, are formed 29.38 grams of N₂O₅.

I hope it helps you!