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

Answer:

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:

$n_{N_{2}} = \frac{m}{M} = \frac{18.5 g}{28.014 g/mol} = 0.66 moles$

$n_{O_{2}} = \frac{m}{M} = \frac{26.7 g}{31.99 g/mol} = 0.84 moles$                                                

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

$n_{N_{2}} = \frac{2}{5}*0.84 moles = 0.34 moles$

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:

$n_{N_{2}O_{5}} = 0.84 moles*\frac{2}{5} = 0.34 moles$  

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:

$m = m_{T}*\frac{80}{100} = 36.72*0.8 = 29.38 g$

Therefore, are formed 29.38 grams of N₂O₅.

I hope it helps you!