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Problem: Metallic aluminum reacts with MnO2 at elevated temperatures to form manganese metal and aluminum oxide. A mixture of the two reactants is 67.2% mole percent Al.Find the theoretical yield (in grams) of manganese from the reaction of 230 g of this mixture.

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FREE Expert Solution

Balanced equation: 4Al + 3MnO2 Ć  3Mn + 2Al2O3


χAl=mole of Almole of mixture


χAl+χMnO2=1


For 1 mole of mixture:


mole of Al=0.672(1 mole mixture)=0.672 mol Al


mole of MnO2=(1-0.672)(1 mole mixture)=0.328 mol MnO2


limiting reactant ā†’ forms less amount of product ā†’ determines theoretical yield


0.672 mol Al×3 mol Mn4 mol Al= 0.504 mol Mn


0.328 mol MnO2×3 mol Mn3 mol  MnO2= 0.328 mol Mn 


limiting reactant: MnO2 (forms less Mn)

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Problem Details

Metallic aluminum reacts with MnO2 at elevated temperatures to form manganese metal and aluminum oxide. A mixture of the two reactants is 67.2% mole percent Al.

Find the theoretical yield (in grams) of manganese from the reaction of 230 g of this mixture.

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Limiting Reagent concept. If you need more Limiting Reagent practice, you can also practice Limiting Reagent practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Porter's class at IUPUI.