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:
[tex] n_{N_{2}} = \frac{m}{M} = \frac{18.5 g}{28.014 g/mol} = 0.66 moles [/tex]
[tex] n_{O_{2}} = \frac{m}{M} = \frac{26.7 g}{31.99 g/mol} = 0.84 moles [/tex]
Now, we need to find the limiting reactant. From reaction (1) we have that 2 moles of N₂ react with 5 moles of O₂:
[tex] n_{N_{2}} = \frac{2}{5}*0.84 moles = 0.34 moles [/tex]
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:
[tex] n_{N_{2}O_{5}} = 0.84 moles*\frac{2}{5} = 0.34 moles [/tex]
Now, we can calculate the theoretical mass of N₂O₅:
[tex] m_{T} = n_{N_{2}O_{5}}*M = 0.34 moles*108.01 g/mol = 36.72 g [/tex]
Finally, if the reaction is only 80% efficient then the mass of N₂O₅ produced is:
[tex] m = m_{T}*\frac{80}{100} = 36.72*0.8 = 29.38 g [/tex]
Therefore, are formed 29.38 grams of N₂O₅.
I hope it helps you!
