If this reaction produced 80.1 g KCl, how many grams of O2 were produced?

Question

When heated, KClO3 decomposes into KCl and O2.

2KClO3 -> 2KCl + 3O2

If this reaction produced 80.1g KCl, how many grams of O2 were produced?

(I don't know if pictures are available to be uploaded, if possible, would you please include a picture? I just tend to have a hard time understanding the math when it's written on a computer in a straight line type thing)

Answer 1

I can't upload a picture but I'll try to explain the thought process and the math as best I can.

First thing we must do is look at the correctly balanced equation for this reaction. It is...

2KClO₃ -> 2KCl + 3O₂

What does this tell us? It tells us that TWO moles of KClO3 decompose to make THREE moles of O2 and TWO moles of KCl. So, if we know the MOLES (not grams) of KCl produced, we can find the MOLES of O2 produced. The O2 will be produced in a ratio of 3/2 relative to KCl. That is the mole ratio obtained from the balanced equation. That is to say, you will get 1.5 times as much O2 as you get KCl (3/2).

But we always work in moles when looking at a balanced equation, and this problem is asking us for grams. It is a simple matter to convert moles of anything to grams, if you know the molar mass. The molar mass of KCl is 74.6 g/mol. The molar mass of O2 = 32 g/mol.

So, to answer the question, we convert 80.1 g KCl to moles of KCl, and then convert moles of KCl to moles of O2 and finally, convert moles of O2 to grams of O2.

moles KCl = 80.1 g x 1 mol KCl/74.6 g = 1.07 moles KCl produced

moles O2 = 1.07 moles KCl x 3 moles O2/2 moles KCl = 1.61 moles O2 produced

grams O2 = 1.61 moles O2 x 32 g O2/mol O2 = 51.5 g O2 produced

I hope this made some sense.