Limiting Reactant Problem

  • Thread starter Chandasouk
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  • #1
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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.
 

Answers and Replies

  • #2
symbolipoint
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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.
 
  • #3
chemisttree
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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.
 
  • #4
Borek
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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.

--
methods
 

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