Potassium chlorate decomposes Into potassium chloride and axygen How manY grams oxygen can produced from 12 Tolc cnlorate? (enter number onty;no: unlc round corert ...

How many grams of potassium chlorate decompose to potassium chloride and 638 $\mathrm{mL}$ of $\mathrm{O}_{2}$ at $128^{\circ} \mathrm{C}$ and 752 torr?
$$2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$

Hi there. Welcome to problem number 44. This is what we call a mass to mass stroke geometry problem because we are given mass of one substance in the equation and we are asked to find the mass of the other. So whenever we have a story geometry problem, we need a balanced equation and that is given to us. We have to KCL- 03 decomposes To Form two KCL and three oxygen. And I left off the states of matter because those are not actually needed to solve this problem. All right. So, we want to know how many grams of potassium chlorine must decompose. So we're trying to find grams of potassium glory to form five grams of the oxygen. All right. So, we're trying to go from massive oxygen. Remember we can start with any reactant or any product in ST geometry. So we're trying to go from grams of oxygen two g of case yellow three. So we will need to use three steps to do this first. We'll go from grams of oxygen. Two moles of oxygen. Using oxygen, solar mass. Then we'll use the mole ratio to go from moles of oxygen. Two moles of case yellow three. And then we will multiply by KCL- 03 smaller masks to give us two g. So that's our general plan, let's go ahead and look at that in each step, adding the numbers. So we're starting with five g of 02. First thing we need to do is to convert two moles of oxygen. So looking up the molar mass of 02, we find that two times 16 point oh oh is 32 point oh oh, So 32g of co two In every model of 02. Next we need to use the mole ratio from the balanced equation and that comes from the coefficients In front of oxygen. There is a three. So for every three moles of 02 And I knew that I wanted to put moles of 02 in the denominator so that it will cancel. We already have grams of oxygen cancelled. All right. So in the new writer will go the moles of case yellow three. And in the balanced equation the coefficient in front of case yellow three is it too? If I were to solve this right now I would have moles of case yellow three. But we are interested in grams. So now I need the molar mass of KCL 03. Remember you can get that by adding together. Okay, SCL and three oxygen's And that ends up being 122.55 g of K C L oh three Notice moles of case yellow three cancel. So we are ready to solve this and our answer will give us g of KCL- 03. We need to round our answer to three significant figures since our initial measurement has three and we're multiplying and dividing. So I get an answer when I calculate this of 12.8 grams of k c l oh three, that is how much we would need to decompose To produce five g of oxygen gas. Hey, thanks so much for watching. I hope this was helpful.

So let's go ahead and start this problem by writing are balanced equation. Okay, we've got some potassium chlorine which is a solid and it's going to produce potassium chloride. She's also a solid and we heated it to do this and the other product is some oxygen gas. So we'll take a minute and go ahead and balance this. So I need a three and a two here to balance the ohs And then a 2 to finish balancing there. So we're going to try and figure out our theoretical yield and we know our percent yield. Okay, so percent yield equals the actual or experimental yield. Whatever you made in the lab, divided by the theoretical yield, what you should have made times 100. So we'll plug in what we know, you know, it was 83 0.2%. Our actual yield was 1985 g of oxygen. That's over our theoretical yield times 100. So our theoretical yield here is going to come out to be 238 .6 g of co two. Uh huh. So we can now use that to figure out the grams of K. C. L. We would need, Okay so I'm gonna say 238 six g of 02 So I'm gonna change g of co two, two moles of 0. 2. Then I can do my more ratio. I can change more than 02.2 over the malls. What I'm looking for which is R. K. C. L. 03 And then finally I can change those moles to grams. So that's 122 0.55 21 Our mole ratio is 2-3 And oxygen which is diatonic is 32. So I can figure out I would have needed 609 g of my potassium chlorate for this to happen at that percentage.

For the balanced equation, we're told it's potassium chlorine. K C. L. 03 turns into potassium chloride, K. C. L. And oxygen gas which is 02 So we'll put it to in front here in a three in front here to balance the oxygen's. And then we needed to in front of the K. C. L. To balance the potassium and chlorine. That's our balance equation. Now we're told that we have 195.8 g. Mhm of 02 that's actually produced. And the yield is um 83.2. So if we divide by that as a decimal .832 we'll get the theoretical yield and then we can do this archaeology. So 1 95.8 divided by .832 is a theoretical yield of 235.3 grams of. Okay. And then we want um the to convert the potassium chlorine So we'll convert two moles. one mole of co two is 32 g. Then we'll use them all ratio 2-3. two moles of case yellow, three. Okay. 23 moles of 02. And then I'll bring it to the next line and when we multiply by the molar mass of potassium chlorate, which is yeah, 122.55. Okay. Okay. Okay. Okay. And that will give us our answer. So we have to 35.3 divided by 32 Times two divided by three And then times 122.55. And we get um 600.8 to 4 significant figures grams of k c l 03 Okay.

Okay, let's calculate how many leaders of oxygen will form from the decomposition. Potassium chlorate? That's right. The decomposition reaction to K C. L 03 to K C. L and three to. So here's their balanced chemical equation. Starting from 20.8 g of potassium clarets convert two moles. Solar masses 122 55 g k. C l 03 in one more K C l 03 start geometry and the mole ratio here. Two moles of K C L 03 to 3 moles of 02 and we're at, uh, STP. So one mole of O to support it to 22.4 leaders of 02 This will work out to 5.70 liters of oxygen produced upon the decomposition of 20.8 g of potassium chlorine