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cody331

Ok, so my teacher gave us a worksheet today Called "calculate the cake"And its pretty much figurein out how many moles and stuff are gonna be left after the reaction.

I listned to the listen, and took notes and all, but im still dumbstrucked by this.

One of the questions on their is this

3. Consider The Following reaction: 2 CuCL2 + 4 KI > 2CuI + 4KCI + I2
a.When 0.56 moles of CuCL2 reacts with 0.64 moles of KI, how many moles of I2 are formed?
b. What is the limiting reactant?
c.For the reactant in excess, how many moles are left over at the end of the reaction?

If somebody could possibly answer, and give a good explanation on how/she answered this, this would greatly, greatly help me, as I love science, and chemistry and all, but some stuff confuses me, and i want to get better.

Thanks in Advance

Eriond

3. Consider The Following reaction: 2 CuCL2 + 4 KI > 2CuI + 4KCI + I2

a.When 0.56 moles of CuCL2 reacts with 0.64 moles of KI, how many moles of I2 are formed?
The reaction uses all 0.64 moles of KI and 0.32 moles of CuCL2 (as you need twice as much KI as you do CuCL2). It forms 0.16 moles of I2, half again as little CuCL2 you need.

b. What is the limiting reactant?
KI. When you reactor 0.64 of KI, you still have 0.56 - 0.32 = 0.24 moles of CuCL2 left. So if you have more KI you could use up all the CuCL2.

c.For the reactant in excess, how many moles are left over at the end of the reaction?
Oops. I did that above. 0.24 moles.

Hope that helps. If you still don't understand, don't hesistate to post.

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MerkaMerka

Why don't you do your own homework? seriously, I remember going over Moles in my Chemistry class(forgotten a good percentage of it) but all it is is algebra, do it yourself, you'll learn more.

^Admin Above Me^

:<...

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Eriond

Lol, sorry. I sort of agree. But I hate it when I just have no clue what's going on in class, and would have loved an explanation the night before the work was due

Besides, this is only one of the problems. He can use this as a starting point.

Also, I'm not an admin Just a mod. Not even a mod of this section. Don't worry, I can't touch you. Not really.

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MerkaMerka

I'm not scared of mods ...

And when I was getting a D in Honors Algebra I had gotten my sister to help me with everything.... and I came out of the class with a high A. Should ask the people around you for help if anything, instead of the internet, since it is quite misleading.

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cody331

I live in hempstead texas, a really low end city, my brother is in the air force and practially almost failed chemistry, my parents cant help b/c they "cant" remember

Tutoring doesent help b/c they monolouge about the same thing and its no help, I know the internet can be misleading, but i put a lil trust in this fourm board since ive been here for about 2 years (registered last year) So thats why i asked

EDIT: And Thanks Eriond
And as for the problem, i kinda get the rest except part A.. =/

MrWafflez

To simplify his answer for better comprehension:

In a balanced equation, there is a ratio of reactant to reactant and reactant to product for each reactant. In this equation, CuCl2 and KI are the reactants, and 2CuI, 4KCI, and I2 are the products.

The ratio of CuCl2 to KI is 1:2, thus for every mole of CuCl2 reacted you also need two moles of KI. In this equation, since you don't have enough KI to react in a 1:2 ratio with the 0.56 moles of CuCl2, it is the limiting reactant.

Then, the ratio of CuCl2 to I2 is 2:1 (Once again, you find the ratio by the coefficients), so divide the moles of CuCl2 consumed in the reaction by 2 to get the moles of I2 produced.

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Eriond

K. I'll try to go through it. First, you have to identify the limiting reagent. It's the chemical that'll react totally in the equation. To find out which one it is, divide the amount of moles of a substance by the integer infront of it in the reaction.

Whichever one is smaller, in this case KI (0.64 / 4 < 0.56 / 2) is the limiting reagent.

Next, you react that in the equation. So you use all of your KI, (.64 moles). Then you divide that amount by the integer infront of it, again, and then multiply it by the integer infront of whatever substance you want to find the amount of moles of, in this case CuCL2, so 2 (the number infront of the CuCL2) * (.64 / 4 (the number infront of KI)) = 2 * (0.64 / 4) = 0.32 moles.

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