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Solld phosphorusand chlorlneqas react t0 fomm solid phosphorus pentachloride (PCL;) Suppose you have 0 mol of P and 2.0 mol of Clz In areactorCalculate the largest ...

Question

Solld phosphorusand chlorlneqas react t0 fomm solid phosphorus pentachloride (PCL;) Suppose you have 0 mol of P and 2.0 mol of Clz In areactorCalculate the largest amount of PCI = that could be produced_ Round your answer to the nearest 0.1 mol.molLro

Solld phosphorus and chlorlne qas react t0 fomm solid phosphorus pentachloride (PCL;) Suppose you have 0 mol of P and 2.0 mol of Clz In a reactor Calculate the largest amount of PCI = that could be produced_ Round your answer to the nearest 0.1 mol. mol Lro



Answers

How many grams of phosphorus react with 35.5 $\mathrm{L}$ of $\mathrm{O}_{2}$ at STP to form tetraphosphorus decoxide?
$$\mathrm{P}_{4}(s)+5 \mathrm{O}_{2}(g) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)$$

Hello students in this question we have given the volume of the hospitals where for equals to 34.05 million liter. And wait, it is equal to w it is 0.625 g at temperature t equals 25 46 degrees integrate. This is equal to 5 46 degrees integrate in the Calvin it is equals 2819 Calvin. Okay. And pressure equals to 0.1 bar. So we have to determine its molecular mass. Okay, so from the ideal situation we can write that PV equals too N. RT. Where this is the number of modes which can be written as given mass developed by molecular mass manipulated by RT. So molecular mass from here can be written as W. R. T. There will be pressure team or player B volume. Okay so we can substitute the value so we get W. Which is equal to Which is equal to 34.05 milli liter. That is the volume. Sorry, W. Which is equals to this 0.625 manipulated by our which is equals two, which is equal to the 82.1 particular bait temperature, which is 819 Calvin developed by pressure P, which is equals two Pressure P, which is equal to 0.1 bar or club Volume V. Which is equal to 34.05. Maybe later. Okay, so from here after solving the molecular mass M, this is equal to 1 24.75 g per board. So this becomes the answer for this problem. Okay, thank you.

So the equilibrium reaction is PCL three for cl two is in equilibrium with PCL five and they're all gases. So our Casey is the concentration of the PCL five provided by the concentration of the PCL three Times The Concentration of the Cl two. So the concentration of our of our PCL three, he's gonna be moles over leaders to a point 0148 moles And we're in four L. So that gives us .003 70 more. The concentration of our PCL five is a .0126 moles By that by four L. So .00315 Moller. And then our concentration of our chlorine, This .08 70 moles in four liters. So 0- 18 Mueller. So we'll go ahead and plug those into our K. Expression or Casey expression. So that's going to be our PCL five, which is 00 315. Try to buy our PCL three our chlorine. So that's going to give us a K. C. Of 39 0.1

So let's start by writing our equation for our equilibrium. So PCL five forms PCL three post chlorine. These are all gases. We'll set up an icebox initial change equilibrium. And we're starting with 2.5 malls and none of these. And we're told that at equilibrium we've got points 338 moles of our PCL three. So we must have added .338. So we'll add .338 of this as well and use up .338. It's all a 1-1 ratio there. So this will give us 2.162. This will give us .338. So we have .338 moles of PCL three and .338 moles of cl two. And we have 2162 moles of our PCL five at equilibrium.

Okay, So in this question, they want us to calculate the maximum amount of P. Ford. I can be produced from one kilogram of fast, right? And then they say the faster I sample contained 75%. See a three parentheses, peel for parentheses to So how much see Ace Threepio for to do we have? Well, they say that we have one kilogram of foster. I've only 75% of that is See a three peel for two. So you multiply that by 75 divided by 100. Then we should get the amount of C A three p o for two. Right? So we get 2.75 kilograms. Let's convert this kilograms in two grams. So one kilogram is equal to 1000 grams and we get 750 grams of C A three p 04 to Okay, So the mole ratio of see a three p o for two is to to one, right. It's okay. So let's start setting up an expression that we can plug into calculated. So we have 7750 grams of C a three Pierre for two. Now we have to convert this into most first. So one mo off see a Threepio four is equal to it's equal to 310 0.18 grams promo. Right. So in order to figure out the molar mass, ivory figured it out before. But if you wanted to do it by yourself, are you gonna do is we have three Most of calcium Multiply it by the molar mass of calcium. Plus we have to most of potassium, right, because to outside the parentheses, that means we must fly everything in the parentheses by two to most of potassium times by the molar mass of potassium plus two times four. So we have eight moles of oxygen times by the molar mass of oxygen, and then you can figure out the mullahs mask. So in order to safe time, I've already calculated of the molar mass. So one So so far, we converted the 750 grams of C A threepio for two into most. So now let's look at the multiple ratio of this and this because this is the product. We're trying to see how much you can produce. So we have two moles of see a three I feel for two produces one mo of P four, right? One mo of P four is equal to 123.88 grams promo and I was gonna do now is just plunges into the calculator and we should get got 1 49.77 grams. If they want you to around a whole number, I guess it can be about 150 grams.


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