About the chemistry calculation

Question

1 moles of KClO3 = 122.55 grams

1 moles of KCl = 74.6 grams

1 moles of O2 = 31.9988 grams

Mass of test tube + MnO2 -> 21.5403g

Mass of test tube + MnO2 + KClO3 + KCl -> 22.5518g

Constant mass of test tube + reaction product -> 22.3514g

1)Mass of O2 released ?

my answer is 22.5518 - 22.3514 = 0.2004g O2

2)Moles of O2 released ?

0.2004g O2 / 31.998g = 0.006263 mol O2

my calculation is correct until now?

I don't know how to solve the questions below?

what is the formula?

3)Moles of KClO3 required by the moles of O2 released ?

4)Mass of KClO3 required ?

5)Mass of KClO3 & KCI mixture?

6)Percent of KClO3 in KClO3 & KCl mixture?

7)If there were 0.5011g KClO3 in the original mixture what is the true percent of KClO3?

Answer 1

Balance the equation, first. It would be 2 KClO₃₌ 2KCl + 3 O_{2 . Number of moles of} O₂ is 3/2 or 1.5 of number of moles of KClO₃: it would be .0093945 moles.

Multiply this answe by the molar mass of KClO₃, 122.55g/mol and your answer is 1.151 g.

Mass of KCLO3 +kcl is 22.5518 g-21.5403g= 1.0115 g

It looks like the mass of KClO3 in your mixture is larger than the mass of the mixture, which is impossible. If the mass of your mixture is bigger, divide the mass of KClO3 by the mass of your mixture to get a % of KClO3.

It's not clear if your data are received experimentally or not. If yes, it means you made a mistake in weighing your samples.

Or maybe I made a technical mistake in my calculations.