Question

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***

Answer 1

29.38 grams

Step-by-step explanation:

The reaction is:

2N₂(g) + 5O₂(g) → 2N₂O₅ (1)

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

`n_{N_(2)} = (m)/(M) = (18.5 g)/(28.014 g/mol) = 0.66 moles`

`n_{O_(2)} = (m)/(M) = (26.7 g)/(31.99 g/mol) = 0.84 moles`

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

`n_{N_(2)} = (2)/(5)*0.84 moles = 0.34 moles`

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:

`n_{N_(2)O_(5)} = 0.84 moles*(2)/(5) = 0.34 moles`

Now, we can calculate the theoretical mass of N₂O₅:

`m_(T) = n_{N_(2)O_(5)}*M = 0.34 moles*108.01 g/mol = 36.72 g`

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

`m = m_(T)*(80)/(100) = 36.72*0.8 = 29.38 g`

Therefore, are formed 29.38 grams of N₂O₅.

I hope it helps you!