Question

27. How many moles of nitrogen are formed when 25.0 g of KNO3 decomposes according to the following reaction? The molar mass of KNO3 is 101.1 g/mol. How many moles of oxygen are formed? Potassium?

4KNO3(s) → 2K2O(s) + 2N2(g) + 5O2(g)

Also, if someone could explain how to obtain the conversion factor, that'd be great. I understand the process of solving a problem like this goes: mass → molar mass → stoich → molar mass → mass, correct?

I'd really like to become an expert with basic stoichiometry problems like this, so if someone could recommend a resource or a great way to practice them, that'd be great too!

Thank you!

Answer 1

4KNO3(s) → 2K2O(s) + 2N2(g) + 5O2(g)

4 moles of KNO3 decomposition to form 2 moles of N2

4*101.1gm of KNO3 decomposition to form 2*28gm of N2

25gm of KNO3 decomposition to form = 2*28*25/4*101.1

                                                             = 1400/404.4   = 3.46gm of N2