My calculation is correct? (about the chemistry)

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

Sorry, I've already submitted the answer, but then realized that I made a mistake. Based on coefficients in the balanced equation, you need to multiply the calculated number of moles of oxygen by 2/3: 0.006263x2/3=.004175 mol KClO3 in the mixture with KCl. Convert to grams by x 122.55 g/mol=.51169 g

% KClO3 =.51169/ mass of the mixture= (.51169g/1.0115g)×100 = 50.587%