How Many Moles of Copper Are Produced from 4.8 Moles of Aluminum?

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Discussion Overview

The discussion revolves around a stoichiometry problem involving the reaction between aluminum and copper(II) chloride to determine the moles of copper produced from a given amount of aluminum. It includes calculations related to both the production of copper from aluminum and the amount of aluminum needed to produce a specific mass of copper.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant initially calculated the moles of copper produced from 4.8 moles of aluminum using the stoichiometric coefficients, arriving at 7.2 moles of copper, but later questioned this due to the limitation of copper available in the reaction.
  • Another participant pointed out that the coefficients in the balanced equation indicate the ratios of reactants and products, not the absolute amounts present.
  • A participant expressed uncertainty about the interpretation of the coefficients, specifically regarding the amount of copper present in the reaction.
  • Calculations were presented for determining the grams of aluminum required to produce 12.7 grams of copper, leading to a conclusion of needing 3.6 grams of aluminum.

Areas of Agreement / Disagreement

Participants generally agree on the interpretation of coefficients in the balanced equation as representing ratios, but there is some uncertainty regarding the initial calculations and assumptions about the amounts of copper produced.

Contextual Notes

There is a lack of clarity on the interpretation of the coefficients in terms of the actual amounts of reactants and products, and the calculations rely on the assumption that the reaction goes to completion.

Jimbo57
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Homework Statement


3CuCl2(aq) + 2Al(s) --> 2AlCl3(aq) + 3Cu(s)

a) If 4.8 mol of Al react in the process, how many moles of Cu are produced?
b) How many grams of Al will be necessary in order to produce 12.7g of Copper?

Homework Equations





The Attempt at a Solution


a) This one is tricky in that I originally did 4.8/2Al X 3Cu = 7.2mol of Cu but then I realized that there is still only 3mol of Cu present as a reactant regardless of the concentration of Al. So I would assume the 3Cu would stay the same.

b) 12.7/63.5 gr Cu= 0.2mol Cu
0.2mol Cu x 2Al/3Cu = 0.13mol Al
0.13mol Al x 27gr/mol = 3.6gr Al for 12.7gr of Cu

How does that look?
 
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Jimbo57 said:
This one is tricky in that I originally did 4.8/2Al X 3Cu = 7.2mol of Cu but then I realized that there is still only 3mol of Cu present as a reactant

Where is it stated?

b) 12.7/63.5 gr Cu= 0.2mol Cu
0.2mol Cu x 2Al/3Cu = 0.13mol Al
0.13mol Al x 27gr/mol = 3.6gr Al for 12.7gr of Cu

How does that look?

Much better.
 
Hey Borek,

I was assuming that 3CuCl2(aq) meant there was 3 mol copper present, but I'm guessing that isn't the case (I'm very new to this). So my original assumption is the better one?
4.8/2Al X 3Cu = 7.2mol of Cu
 
Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.
 
Borek said:
Yes. Coefficients don't say anything about the amounts - only about their ratios during reaction.

Thanks for the help :)
 

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