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.

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.

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

Well, let's do some math to find the theoretical yield of manganese in this reaction.

First, we need to find the moles of aluminum present in the mixture. Since the mixture is 67.2% mole percent Al and we have 240 g of the mixture, we can calculate the moles of Al using the molar mass of Al (26.98 g/mol):

(240 g) * (0.672) / (26.98 g/mol) = 5.954 moles of aluminum

According to the balanced equation, the reaction between aluminum and manganese dioxide has a 3:1 stoichiometric ratio, meaning that for every 3 moles of aluminum reacting, 1 mole of manganese is produced.

So, the moles of manganese produced can be calculated as:

5.954 moles of Al * (1 mole of Mn / 3 moles of Al) = 1.985 moles of manganese

Now, to find the molar mass of manganese (54.94 g/mol) and the theoretical yield of manganese:

(1.985 moles of Mn) * (54.94 g/mol) = 109.04 grams of manganese

Therefore, the theoretical yield of manganese from 240 g of the mixture is approximately 109.04 grams. Remember, this is theoretical, assuming 100% yield and no losses during the reaction.

To find the theoretical yield of manganese (Mn) from the reaction, we need to follow these steps:

Step 1: Convert the given mass of the mixture to moles.

To do this, we divide the mass of the mixture by its molar mass. The molar mass of aluminum oxide (Al2O3) is 101.96 g/mol.

240 g * (1 mol / 101.96 g) = 2.354 mol

Step 2: Determine the moles of aluminum (Al) in the mixture.

Since the mixture is 67.2% mole percent Al, we can calculate the moles of Al using the following expression:

moles of Al = 0.672 * total moles of mixture
moles of Al = 0.672 * 2.354 mol
moles of Al = 1.579 mol

Step 3: Determine the stoichiometry of the reaction.

Balanced chemical equation:
2Al + MnO2 -> Mn + Al2O3

From the equation, we can see that 2 moles of Al react with 1 mole of MnO2 to produce 1 mole of Mn.

Step 4: Calculate the moles of Mn produced.

Since the ratio of Al to Mn is 2:1, the number of moles of Mn produced will be the same as the moles of Al, which is 1.579 mol.

Step 5: Calculate the molar mass of Mn.

The molar mass of manganese (Mn) is 54.94 g/mol.

Step 6: Calculate the theoretical yield of Mn in grams.

theoretical yield (grams) = moles of Mn * molar mass of Mn
theoretical yield = 1.579 mol * 54.94 g/mol
theoretical yield = 86.692 g

Therefore, the theoretical yield of manganese (Mn) from the reaction of 240 g of the mixture is 86.692 grams.

zubi