Stoichiometry - Reactants & Product Masses

[markelmarcel]

Homework Statement

A student has 4.00g of Copper Chloride (CuCl2) and 25.0 g of Aluminum (Al). Calculate the mass of each reactant and product present at the end of the reaction. Show your calculations.

Homework Equations

3 CuCl2 + 2Al $$\rightarrow$$ 3Cu + 2AlCl3

The Attempt at a Solution

Ok. I already know that Copper Chloride is the limiting reactant, but I know how to prove it too. I have figured out the amount of Copper that would be produced from 4.00g of CuCl2 as 1.89g Cu.

I have figured out that the amount of Copper that would be produced from 25.0 g of Al as 88.3g Cu.

So, like I said... CuCl2 is the limiting reactant.

I also know that only 0.535g Al reacts with the 4.00 g of CuCl2.

But- I honestly don't get what this question is actually asking me to do... it's driving me nuts.

 

[symbolipoint]

You say you know that the CuCl₂ is the limiting reactant. This means that ALL of it will react. This also means that a portion of the Al will reactant, therefore the other portion of Al will remain unreacted.

 

[markelmarcel]

You say you know that the CuCl₂ is the limiting reactant. This means that ALL of it will react. This also means that a portion of the Al will reactant, therefore the other portion of Al will remain unreacted.

Correct. So- I would know that all 4.0g of CuCl₂ will react... what I don't know is how much of the 25.0g Al will react with that.

so I set up 4.0g CuCl₂ changed it into moles (1 mol/134.45 g CuCl₂ )

Then from moles of CuCl₂ I can flip over to moles Al using the balanced equation ( 2 mol Al / 3 mol CuCl₂ )

Then from moles Al, I convert to g Al... (1 mol Al / 26.98g Al)

And I got... 0.535g Al

So then I just take 25.0g Al - 0.535g Al = 24.465g Al as the portion being unreacted.

And then, is that it? I've solved everything that they have asked for??

 

[symbolipoint]

Your process looks good; I did not examine it for the calculational results. You also use 'stoichiometry' to determine the amounts of products.

 

[Borek]

And then, is that it? I've solved everything that they have asked for??

No, you have solved for Al. What about other substances present after?

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[markelmarcel]

No, you have solved for Al. What about other substances present after?

How do I do the AlCl₃? Start with the 4.00 g CuCl₂ convert to moles of CuCl₂, then moles of AlCl₃ and then to grams? Will that tell me what is left over?



PS -- Thanks for everyone's replies.

 

[Borek]

How do I do the AlCl₃? Start with the 4.00 g CuCl₂ convert to moles of CuCl₂, then moles of AlCl₃ and then to grams?

Yes. Or go directly from amount of reacting Al that you have already calculated.

Will that tell me what is left over?

No, but you have already calculated how much Al was used in the reaction. You know you started with 25.0 g, you know 0.535g reacted, I hope calculating how much was left is not beyond your comprehension

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[markelmarcel]

I hope calculating how much was left is not beyond your comprehension

Ok! I got all the parts then! I got my Al, my CuCl₂, Cu, and AlCl₃.

And, you know the calculation of how much was left was quite difficult! It took me awhile to figure it out! Haha. Juuust kidding.

Thanks again! :)

 

[Borek]

So, how much CuCl₂?

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