I was told to post at the beginning of the page: not sure if I did so or how to really do it but i'm posting the question again):

Diiodine pentafluoride reacts spectacularly with bromine trifluoride to form iodine pentafluoride, oxygen gas, and liquid bromine. In a particular reaction a 44.00g sample of I2O5 reacts with 58.00g BrF3. What is the mass of the excess reagent?

This is what I have so far:
6I2O5 + 20BrF3--> 12IF5 + 15O2 +10Br2

44gI2O5 (1mol/333.795)=0.1318 mol I2O5

58.00gBrF3 (1mol/136.898)= 0.1853 mol BrF3

0.1318mol I2O5 (20molBrF2/6 mol I2O5)= 0.4393 mol BrF3

0.1853 mol BrF3(6 mol I2O5/20mol BrF3)=0.0556mol I2O5

I believe you missed the point of my response. My response was that the first sentence waxes supreme about diiodine pentafluoride reacting spectacularly with BrF3 BUT YOUR EQUATION USES I2O5 AND IF5 is a PRODUCT and not a reactant.

I assume you know this is a limiting reagent problem because the question is how much EXCESS reagent remains (I assume the problem means when the reaction is over). Here is what you posted.
This is what I have so far:
6I2O5 + 20BrF3--> 12IF5 + 15O2 +10Br2

44gI2O5 (1mol/333.795)=0.1318 mol I2O5 This step is right.

58.00gBrF3 (1mol/136.898)= 0.1853 mol BrF3
This step is done right but you must have punched in the wrong key because I obtained 0.4237

0.1318mol I2O5 (20molBrF2/6 mol I2O5)= 0.4393 mol BrF3
I think you have assumed here that I2O5 is the limiting reagent and that's isn't true.

0.1853 mol BrF3(6 mol I2O5/20mol BrF3)=0.0556mol I2O5
<b?Again, I think you assumption on the limiting reagent is not right so look below for finishing the problem.

Here is what we have from above.
mols I2O5 = 0.1318
mols BrF3 = 0.4237

How much Br2 (nothing says we must use Br2--we could use any product--I just picked Br2) would be formed if we used all of the I2O5 and had all of the BrF3 we needed. That is
0.1318 x (10 mols Br2/6 mols I2O5) = 0.2197

How much Br2 would be formed if we used all of the BrF3 and had all of the I2O5 we needed. That is
0.4237 x ((10 mols Br2/20 mols BrF3) = 0.2119.

You see the two answers don't agree, which is usual for limiting reagent (LR) problems which means one of them is not right. The correct value in LR problems is ALWAYS the smaller number and the reagent producing that value is the LR. So BrF3 is the limiting reagent.

We can now calculate how much Br2 is formed (the problem doesn't ask for that) and it is we form 0.2119 mols Br2.

Now to the question of how much of the excess reagent (that's the I2O5) remains after reacting. We must find how much is used.
That's 0.4237 mols BrF3 x (6 mols I2O5/20 mols BrF3) = 0.1271 mols I2O5 used. We had 0.1318 initially; therefore, the amount that is left is
0.1318-0.1271 = ? mols I2O5 remains un-reacted. Then mols x molar mass = grams.

To find the mass of the excess reagent, we need to determine which reactant is limiting and which is in excess.

First, we will calculate the theoretical amount of each reactant in the given reaction:

From the reaction equation: 6I2O5 + 20BrF3 → 12IF5 + 15O2 + 10Br2

The molar mass of I2O5 is 333.795 g/mol, and the given mass of I2O5 is 44.00 g.
44.00 g I2O5 * (1 mol I2O5 / 333.795 g I2O5) = 0.1318 mol I2O5

The molar mass of BrF3 is 136.898 g/mol, and the given mass of BrF3 is 58.00 g.
58.00 g BrF3 * (1 mol BrF3 / 136.898 g BrF3) = 0.1853 mol BrF3

Now, we need to determine which reactant is limiting:

The balanced chemical equation tells us that 6 moles of I2O5 react with 20 moles of BrF3.

0.1318 mol I2O5 * (20 mol BrF3 / 6 mol I2O5) = 0.4393 mol BrF3 required (based on I2O5)

0.1853 mol BrF3 * (6 mol I2O5 / 20 mol BrF3) = 0.0556 mol I2O5 required (based on BrF3)

Since 0.1318 mol I2O5 is less than 0.4393 mol BrF3, I2O5 is the limiting reactant. This means that BrF3 is in excess.

To find the mass of excess BrF3:
The molar mass of BrF3 is 136.898 g/mol.
0.0556 mol BrF3 * (136.898 g BrF3 / 1 mol BrF3) = 7.60 g BrF3

Therefore, the mass of the excess reagent (BrF3) is 7.60 g.