Answer:
1. 10.8 g of NO
2. N₂ is the limting reagent
3. 3.2 g of O₂ does not react
Explanation:
We determine the reaction: N₂(g) + O₂(g) → 2NO(g)
We need to determine the limiting reactant, but first we need the moles of each:
5.04 g / 29 g/mol = 0.180 moles N₂
8.98 g / 32 g/mol = 0.280 moles O₂
Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂
Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO
Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 = 0.360 moles NO
If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g
As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.
As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.
0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.
