Chemistry

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 240 g of this mixture.

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  1. One way you can do this, and I don't know that it is the easiest is as follows:
    1. 0.672 x molar mass Al = g Al.
    2. 0.328 x molar mass MnO2 = g MnO2.
    3. Add g Al + g MnO2 to obtain total grams (with the desired mole ratio), then determine the percent Al and percent MnO2 from those values. For example
    g Al = about 18, g MnO2 = about 28, total is about 46 (you need to do these more accurately). So
    % Al = 18/46 = about 0.39 = 39%
    % MnO2 = 28/46 = about 0.61 = 61%

    4. Now multiply the percent by mass x the 240 to find how much Al and how much MnO2 are there.
    240 x 0.39 = about 94 g Al
    240 x 0.61 = about 146 g MnO2.

    5. Knowing 146 g MnO2 means you can get, theoretically,
    146 g MnO2 x (1 mol MnO2/86.93)(54.93 g Mn/1 mol Mn) = ?? g Mn.
    I hope this helps. But there must be a shorter way. Check my work. And redo all of the calculations with more accurate numbers.

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  2. What volume of 0.178 solution is necessary to completely react with 92.3 of 0.106?

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  3. zubi

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