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?

I don't understand

To solve problem b, you need to use stoichiometry, which involves using the balanced chemical equation and the molar ratios to convert between different substances in the reaction. Here's how you can set up the calculation:

1. Start by writing the balanced chemical equation: 2KClO3 → 2KCl + 3O2. This equation shows that for every 2 moles of KClO3, you will get 2 moles of KCl.

2. Convert the given mass of KClO3 into moles. The molar mass of KClO3 is 122.55 g/mol. Therefore, to convert grams to moles, you divide the given mass (53.0 g) by the molar mass:

Moles of KClO3 = 53.0 g / 122.55 g/mol

3. Use the mole ratio from the balanced equation to convert moles of KClO3 to moles of KCl. According to the balanced equation, the mole ratio between KClO3 and KCl is 2:2:

Moles of KCl = Moles of KClO3

4. Finally, convert the moles of KCl to grams. Since the molar mass of KCl is 74.55 g/mol, multiply the moles of KCl by the molar mass:

Grams of KCl = Moles of KCl x Molar mass of KCl

This calculation will give you the answer to problem b, which is the mass of potassium chloride produced from the given mass of potassium chlorate.

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.??

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????