What Determines the Limiting Reactant in a Chemical Reaction?

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Discussion Overview

The discussion revolves around identifying the limiting reactant in a chemical reaction involving aluminum and oxygen, based on a specific mass of each reactant. The context includes a homework problem that requires participants to analyze the stoichiometry of the reaction and determine the mass of the excess reactant after the reaction is complete.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant states that aluminum produces 1.56 mol of Al2O3 and oxygen produces 2.45 mol, concluding that aluminum is the limiting reactant.
  • Another participant questions the mole ratio of aluminum to oxygen and suggests comparing the theoretical mole ratio to determine the limiting reactant.
  • A participant asks for clarification on the number of moles of aluminum and oxygen based on their given masses, indicating a need for foundational calculations.
  • One participant provides a calculation for the moles of aluminum, suggesting that the initial molar calculation for aluminum may contain a typo and confirms the results for oxygen.
  • A suggestion is made to reverse the approach to calculate the amount of Al2O3 from the given amount of O2, emphasizing the use of the same reaction equation and coefficients.

Areas of Agreement / Disagreement

Participants express differing views on the calculations and interpretations of the limiting reactant, indicating that there is no consensus on the correct approach or results at this stage.

Contextual Notes

Some calculations depend on accurate conversions between grams and moles, and there may be unresolved issues regarding the initial calculations and assumptions made about the stoichiometry of the reaction.

Chandasouk
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Homework Statement



A mixture of 82.49g of aluminum and 117.65g of oxygen is allowed to react. Find the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Balanced Equation is 4Al(s) + 3O2 ----> 2Al2O3


I found that Aluminum produces 1.56 mol of 2Al2O3

and Oxygen produces 2.45 mol of 2Al2O3

So, Aluminum is the limiting reactant. I do not know how to find out the grams of Oxygen from here though.


I've been wondering since I could not do this on my first Chem exam.
 
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What is the mole ratio of Al to O2 for the these elements available to react? What is the mole ratio of Al to O2 for the theoretical written reaction? Compare these and decide which element is the limiting reactant. How many moles of the excess reactant are present at completion of the reaction? Convert this back to grams.
 
I don't understand your answer. How many moles of aluminum in 82.49 g? How many moles of O2 in 117.65 g? Start from there.
 
That's (almost) OK - 82.49 g Al -> 3.057 moles of Al -> 1.529 moles of Al2O3 - I suppose 1.56 is just a typo or math error, as results for oxygen are OK.

Chandasouk: you know how to calculate amount of Al2O3 form a given amount of O2, all you have to do is to reverse the approach. Same reaction equaton, same coefficients, just switch kgiven and unknown.

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methods
 

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