Been stumped on this problem for a day now.

2KClO3-->2KCl+3O2

a) If 53.0g of potassium chlorate breaks down, what volume of oxygen is produced?

For this one I got: 14.6L O2

b) How many grams of potassium chloride are produced from problem a?

For the life of me I can't figure out the how much potassium chloride is produce.

How do I set problem b up correctly?

8 answers

Wait a minute here is this the correct answer..I might of been over thinking this one:

Answer: 75g/mol KCl x 2 = 150g KCl Produced????
122.5 grams/mol for KClO3
so we have 53/122.5 = .433 mols of KClO3

for every 2 mols of KClO3 we get 3 mols of O2

(3/2) (.433) = .649 mols of O2
If at STP then about 22.4 liters/mol
so
.649 * 22.4 = 4.5 liters agreed

b).433 mols of KClO3 ---> .433 mols of KCl = 74.5 g/mol
.433 * 74.5

LOOK at the balanced equation for how many mols of this for mols of that
sorry about being anonymous. My computer does that sometimes.
think molecules , do grams later

(molecules/mols is like eggs to dozens of eggs, just 6*10^23 instead of 12)
So for problem a. I have the answer of 14.6L O2 which they gave us so we could learn how to solve it so I know 14.6L O2 is correct unless they gave me wrong answer.

Being that your answer is a bit off from problem a. I'm thinking we may be a bit off on problem b? What do you think?
Hey Damon- Thanks for the tip on the molecules/mols. This will help. I'm so new at this stuff I'm trying to keep up.
For problem a. I did this:

53gKClO3/122gKClO3 * 3molesO2/2molesKCLO3 *
22.4L = 3562/244 = 14.6L O2

That's how I came up with 14.6L O2

Now I'm not sure about problem b. still.

What do you think about for problem b.??
I don't understand
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