# 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 à 3Mn + 2Al2O3

$\overline{){{\mathbf{\chi }}}_{{\mathbf{Al}}}{\mathbf{+}}{{\mathbf{\chi }}}_{{{\mathbf{MnO}}}_{{\mathbf{2}}}}{\mathbf{=}}{\mathbf{1}}}$

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) ###### 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.