Stoichiometry - Reactants & Product Masses

In summary, after completing the reaction, there will be 1.89g of copper, 0.535g of aluminum, and 2.67g of aluminum chloride present. The limiting reactant is CuCl2 and 24.465g of aluminum will remain unreacted. The calculations were completed using stoichiometry and converting between moles and grams.
  • #1
markelmarcel
21
0

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 [tex]\rightarrow[/tex] 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.
 
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  • #2
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.
 
  • #3
symbolipoint said:
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??
 
  • #4
Your process looks good; I did not examine it for the calculational results. You also use 'stoichiometry' to determine the amounts of products.
 
  • #5
markelmarcel said:
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?

--
 
  • #6
Borek said:
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.
 
  • #7
markelmarcel said:
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 :wink:

--
 
  • #8
Borek said:
I hope calculating how much was left is not beyond your comprehension :wink:

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! :)
 
  • #9
So, how much CuCl2?

--
 

1. What is stoichiometry and why is it important in chemistry?

Stoichiometry is the quantitative study of the relationship between reactants and products in a chemical reaction. It is important in chemistry because it allows us to determine the amounts of reactants and products involved in a reaction, as well as predict the outcome of a reaction.

2. How do you calculate the reactant and product masses in a chemical reaction?

To calculate the reactant and product masses, you must first write a balanced chemical equation. Then, using the coefficients in the equation, convert the given mass of one substance to moles. Finally, use the mole ratios from the balanced equation to calculate the masses of the other substances involved.

3. What is the difference between theoretical and actual yield in stoichiometry?

Theoretical yield is the amount of product that is expected to be produced based on the balanced chemical equation. Actual yield is the amount of product that is actually obtained in an experiment. The difference between the two is known as the percent yield, which can be used to assess the efficiency of a reaction.

4. Can you use stoichiometry to determine the limiting reactant in a reaction?

Yes, stoichiometry can be used to determine the limiting reactant in a reaction. This is done by comparing the amount of product that can be produced from each reactant based on the balanced equation. The reactant that produces the least amount of product is the limiting reactant.

5. How does stoichiometry apply to real-world situations?

Stoichiometry is used in many real-world situations, such as in the production of chemicals, pharmaceuticals, and food. It is also used in environmental studies to determine the amounts of pollutants produced in a reaction. Additionally, it is used in everyday life, such as in cooking and baking, to ensure the right amount of ingredients are used to produce a desired outcome.

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