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According to the following reaction: CSz ($) + 4Clz (g) CCh (U) + 2SClz ($)What would you multiply grams of carbon disulfide" by to convert to the units "...

Question

According to the following reaction: CSz ($) + 4Clz (g) CCh (U) + 2SClz ($)What would you multiply grams of carbon disulfide" by to convert to the units "moles of chlorine"(number)(unit)grams CS}(number)(unt)moles Clz(number)(unil)(number)(unit)

According to the following reaction: CSz ($) + 4Clz (g) CCh (U) + 2SClz ($) What would you multiply grams of carbon disulfide" by to convert to the units "moles of chlorine" (number) (unit) grams CS} (number) (unt) moles Clz (number) (unil) (number) (unit)



Answers

The reaction of carbon disulfide with chlorine is as follows:
$$\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)+238 \mathrm{kJ}$$ Predict the effect of the following changes to the system on the direction of equilibrium. (a) The pressure on the system is doubled. (b) $\mathrm{CCl}_{4}$ is removed as it is generated. (c) Heat is added to the system. (d) A catalyst is added to speed up the reaction.

Hi there. Welcome to problem number 87. in problem number 87 we are given 62.7 g of cl two and we want to calculate the number of grams of carbon di sulfide CS two, The Question Mark, g of CS two. So this is a mass to mass problem. Specifically a masked mastoi geometry problem. Because we're starting with grams of Cl two And we're trying to find g of c. S. that's going to react with that. So, to do a strike geometry problem, we know that we're going to need a mole ratio. Well, we have a drawing here. The drawing shows us that for everyone CS two, it looks like we need three of those cl 2s. I was looking at the picture, we see three of those cl to us there. So it's a 1-3 ratio and we will need that here in a little bit. But let's go ahead and get started because what we need to do is convert from g to moles Of the CL two. Then we can convert two moles Of the CS two And then two g of CS two. Right? So those are three steps will be doing here. So let's go ahead and get started. First thing is to convert from 62.7 g of cl two. Jamal's of Cl two. And we're going to do that by using Molar Mass of Cl two. Since it's Cl two, we need to take two times the mass of chlorine, Which gives us 70.90 g. And there are that many g in one mole. And I put the g of cl two in the denominator. So that will cancel. So now we're at moles of cl two. We are ready to use that mole ratio. Remember from the drawing for every three Multiples in this case, M.nolds of Cl 2? We need to react one CS two. All right, so moles of seal to cancel. Now. Finally, we need to convert from moles of CS 2, 2 g. So we need the molar mass of CS two In every mole of CS two. There are 76.15 g. Notice to get from moles to grams were multiplying by this Mueller mass because that way the moles will cancel Right, calculating our answer and rounding it to three significant figures. Since 62.7 had three significant figures, I get 22 0.4 g of CS two is required to react completely With that 62.7 g of chlorine. Right? So 22.4 g of CS two. And that is our answer.

So for this problem, first of all, we're going to work backwards. So we know we have 125 g of the CCL two F two. So the second step gives us a CCL two F two and this is a 71.5% yield. So we need to find the 100% yield for this reaction so that we can find the story geometry correct. So this storytime metric equivalents correctly, so therefore divide by .715 and we get 100% yield to step two. Then we can use this mass divide by the moments to get moles of the CCL two F two. Then use the mole ratio from the second reaction to convert from bowls of CCL two F 22 moles of CCL floor. So therefore we are going to use that mole ratio and now we have moles of C c L four that we obtained her step one. So this is the amount of CCL four that will be required to get the 71.5% yield of CCL four because remember we're only getting a percentage of what we would actually get. So then for the first step we got a 62.5% yield. So this is 62.5% of something. So we have to find out what 100% yield for the first step would be. So divide this number of moles by 0.6 to five. Then we get 100% yield for Stepan now that we have 100% yield, we can find the Storycorps metric equivalent of the c. S. two that will be required to give us this amount of the CCL four. So we're going to use the mole ratio from reaction one to convert from CCO forward to see S two. That's a 1 to 1 mole ratio, then multiplied by the molar mass to get the massive CS two and then report to three significant figures. We get 176 g of CS two.

Doing this problem, we have obtained 125 g of CCL two F two. So let's work backwards. So when CCL two F two is produced when CCL floor from reaction to We got a 71.5% yield. So let's figure out what the 100% you would be. So this is equal 2.715 times X, which is the amount that's 100% yield Because we only got 71.5% of X. So divide what's up a .715 and we get this as a 100% yield. So then we're going to use the balance kind of correction to convert to the mass of the carbon tetrachloride. Yeah, that will give us this yield. So we're going to take 100% yield of this compound divided by the molar mass of the compound. And then from about chemical reaction, three moles of carbon tetra or I'd react to give us one mould this compound. So we're going to use that moderation to convert two moles of carbon tetrachloride, multiplied by the molar mass of carbon tetrachloride to give us the mass of carbon tetrachloride. That will give us this much of the compound. It 100%. You. So now we're told that carbon tetrachloride is produced from carbon di sulfide With the 62 point 5% yield or reaction one. So let's figure out what 100% yield of this would be. So we produce this much and this is how Much reacted in a second reaction. But this is 62.5% of the original value, which is 100% yield. So we're going to perform the same calculation to find 100% yield. So divided by .65 and we get that this is 100% yield, 100% yield from the first reaction. So take this mass divided by the molar mass of carbon tetra chloride. Then we get moles of carbon tetrachloride and then convert two moles of carbon di sulfide and we can see that one mole of carbon disqualified, gives us one mole of carbon tetrachloride, so then multiplied by the molar mass of carbon di sulfide to get the massive carbon disqualified. That will give us this much of the carbon tetrachloride and we get the original mass of the carbon thyself.

To determine the mass of carbon di sulfide that is required For a reaction with 62.7 g chlorine. In the reaction that produces carbon tetrachloride, we need the balanced chemical reaction that produces carbon tetrachloride from carbon di sulfide and chlorine gas. This is provided in your textbook, We will start with a 62-7 g chlorine gas, Convert these g into moles by divided by the Molar Mass of Cl two. Then when we are in units of moles cl two, all we need to do is look at the stock geometric relationship between cl two and the carbon di sulfide. The relationship being one mole. Carbon di sulfide is required for every three moles of cl two. Then, when we have moles carbon di sulfide, we simply multiply by the molar mass. Carbon di sulfide to get grams and we get 22.4 g carbon di sulfide.


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