4 NH3 (g) + 5 O2 (g) → → 4 NO (g) + 6 H2O (l) How many grams of NO is produced if 22.8 grams of NH3 reacts with 55.6 grams of O2?

Answer 1

To solve this problem, you will have to do two separate calculations. Calculate the amount of NO formed by 22.8g NH3 then calculate the amount of NO formed by 55.6g O2. Compare the two answers - the smaller one will be your answer because it is the limiting reagent.

It's a bit hard to type out but I'll give it a try.

1) Calculate amount of NO formed by 22.8g NH3

Convert g NH3 -> mol NH3 22.8g NH3 / 17g/mol NH3 = 1.341 mol NH3

Convert mol NH3 -> mol NO 1.341 mol NH3 = 1.341 mol NO (because they have a 1:1 ratio)

Convert mol NO -> g NO 1.341 mol NO x 30 g/mol NO = 40.2 g NO

2) Calculate amount of NO formed by 55.6g O2

Convert g O2 -> mol O2 55.6g O2 / 32 g/mol O2 = 1.737 mol O2

Convert mol O2 -> mol NO 1.737 mol O2 x (4 mol NO / 5 mol O2) = 1.39 mol NO

Convert mol NO -> g NO 1.39 mol NO x 30 g NO = 41.7 g NO

The amount of NO produced is 40.2 g

Let me know if you have any other questions!