Metallic aluminum reacts with MnO₂ at elevated temperatures to form manganese metal and aluminum oxide.A mixture of these two reactants is 67.2% mole percent Al.Find the theoretical yield (in grams)of manganese from the reaction of 250 g of this mixture.

Question

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First, we need to find the moles of each reactant in the mixture: - Let x be the mole fraction of MnO₂ in the mixture. - Then, the mole fraction of Al is (1 - x). - We know that the mixture is 67.2% mole percent Al, which means: (1 - x) = 0.672 x = 0.328 - This means that for every 1 mole of MnO₂, there are 1.835 moles of Al. Next, we need to use the balanced chemical equation to determine the mole ratio of MnO₂ to Mn: 2 MnO₂ + 5 Al → 2 Mn + Al2O3 The ratio is 2 moles of MnO₂ to 2 moles of Mn, or 1:1. Since we know the mole ratio of MnO₂ to Al, we can set up a proportion to find the moles of Mn: 1 mol MnO₂ / 1.835 mol Al = x mol MnO₂ / (0.328 mol Al) Solving for x gives: x = (1 mol MnO₂ / 1.835 mol Al) * (0.328 mol Al) = 0.179 mol MnO₂ Since the ratio of MnO₂ to Mn is 1:1, the moles of Mn produced by the reaction is also equal to 0.179 mol. Finally, we can convert moles of Mn to grams using the molar mass of Mn: 0.179 mol Mn * (54.94 g / mol) = 9.85 g Mn Therefore, the theoretical yield of Mn from 250 g of the mixture is 9.85 g. Note: Choices A, B, C, and D are all incorrect calculations or conversions.

Metallic Aluminum Reacts with MnO2 at Elevated Temperatures to Form

Metallic Aluminum Reacts with MnO₂ at Elevated Temperatures to Form