If 33.6 g of NO and 26.9 g of O₂ react together, what is the mass in grams of NO₂ that can be formed via the reaction below? 2 NO (g) + O₂ (g) → 2 NO₂ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagent and product participate in the reaction:

NO: 2 moles
O₂: 1 mole
NO₂: 2 moles

Being:

N: 14 g/mole
O: 16 g/mole

the molar mass of the compounds participating in the reaction is:

NO: 14 g/mole + 16 g/mole= 30 g/mole
O₂: 2*16 g/mole= 32g/mole
NO₂: 14 g/mole + 2*16 g/mole= 46 g/mole

Then, by stoichiometry of the reaction, the following amounts of reactant and product participate in the reaction:

NO: 2 moles* 30 g/mole= 60 g
O₂: 1 mole* 32g/mole= 32 g
NO₂: 2 moles* 46 g/mole= 92 g

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

You can apply the following rule of three to determine the limiting reagent: if by stoichiometry 60 g of NO react with 32 g of O₂, 33.6 g of NO with how much mass of O₂ will it react?

[tex]mass of O_{2}=\frac{33.6 grams of NO*32 grams of O_{2} }{60 grams of NO}[/tex]

mass of O₂= 17.92 grams

But 17.92 grams of O₂ are not available, there are 26.9 grams available. Since it has more mass than it needs to react with 33.6 grams of NO, NO will be the limiting reagent.

Then you apply the following rule of three: if by reaction stoichiometry 60 grams of NO form 92 grams of NO₂, 33.6 grams of NO how much mass of NO₂ does it form?

[tex]mass of NO_{2}=\frac{33.6 grams of NO*92 grams of NO_{2} }{60 grams of NO}[/tex]

mass of NO₂= 51.52 grams

If 33.6 g of NO and 26.9 g of O₂ react together, 51.52 grams of NO₂ can be formed.