Carsyn E.

asked • 02/17/24

Solving moles-to-moles

Solid aluminum and oxygengas react to form solid aluminum oxide. Suppose you have 5.0 mol of Al and 9.0 mol of O2 in a reactor.

Suppose as much as possible of the Al reacts. How much will be left? Round your answer to the nearest 0.1 mol.


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Russell G. answered • 02/18/24

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The Al would be completely used because there is excess O2 and there is enough Al to make 2.5 moles of Al2O3.

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J.R. S.

tutor
Correct but one should see the balanced equation to justify the fact that the mole ratio of Al to O2 is 2:3.
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02/18/24

Russell G.

tutor
I approached it slightly differently: 5.0 mol Al x (1 mol Al2O3) / (2 mol Al)=2.5 mol Al2O3 9.0 mol O2 x (2 mol O) / (1 mol O2) x (1 mol Al2O3) / (3 mol O) = 6 mol Al2O3 so there was enough O2 to make 6 moles of Al2O3, but only enough Al to make 2.5 moles of Al2O3.
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02/18/24

Russell G.

tutor
now that I have eaten brekfast, I see what you were getting at. How did I know that it was Al2O3. If I didn't just happen to remember that one, I would have had to balance the equation.
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02/18/24

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