How many grams of NO2 can be formed from the reaction between 28 grams of nitrogen and 64 gram of oxygen

Question

How many grams of NO2 can be formed from the reaction between 28 grams of nitrogen and 64 gram of oxygen

Answer 1

This question has already been asked and answered (TWICE).

Theoretical yield or NO₂ = 1.0 mole N₂ x 2 moles NO₂/mole N₂ x 46 g/mole NO₂ = 92 g NO₂

Answer 2

For this you will have to write a synthesis equation to find the limiting reagent. Since both nitrogen and oxygen are diatomic, the starting compounds will be N₂ and O₂.

N₂+O₂→NO₂ is the unbalanced equation. Now to balance!

N₂+2O₂→ 2NO₂

So now, we can start using stoichiometry. First will begin with 28 grams of nitrogen

28 g N₂* 1 mole/28 g N₂ * 2 NO₂/ 1 N₂ * 46 g NO₂/1 mole= 96 g NO₂

Next, we do the same thing but with 64 g of oxygen.

64 g O₂ * 1 mole/32 g O₂* 2 NO₂/2 O₂ * 46 g NO₂/ 1 mole= 92 g NO₂

So the limiting reagent is oxygen, so the amount of NO₂ will be 92 g.