Asked by Johnny

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?
Answered by Johnny
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????
Answered by Anonymous
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

Answered by damon
sorry about being anonymous. My computer does that sometimes.
Answered by Damon
think molecules , do grams later

(molecules/mols is like eggs to dozens of eggs, just 6*10^23 instead of 12)
Answered by Johnny
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?
Answered by Johnny
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.
Answered by Johnny
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.??
Answered by Jessica
I don't understand
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