Balanced equation: 4Al + 3MnO2 à 3Mn + 2Al2O3
For 1 mole of mixture:
0.672 mol Al
0.328 mol MnO2
limiting reactant → forms less amount of product → determines theoretical yield
= 0.504 mol Mn
= 0.328 mol Mn
limiting reactant: MnO2 (forms less Mn)
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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