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.

Answer 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.

Answer 2

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

Answer 3

zubi