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)= 0.4237 mol BrF3
If you still want to see this done please post again at the top of the page and when you do note that the first sentence talks about diiodine pentafluoride but the reaction you give with the problem uses I2O5 (which I always called iodine pentoxide but I guess the proper name is diiodine pentoxide now.)
To find the mass of the excess reagent, we need to determine which reactant is limiting and which is in excess.
1. Find the molar ratio between I2O5 and BrF3 in the balanced equation:
From the balanced equation: 6 moles of I2O5 react with 20 moles of BrF3.
So, the ratio is: (6 mol I2O5) / (20 mol BrF3) = (3 mol I2O5) / (10 mol BrF3)
2. Calculate the number of moles of BrF3 needed to react with the given amount of I2O5:
Number of moles of BrF3 needed = (0.1318 mol I2O5) * (10 mol BrF3) / (3 mol I2O5) ≈ 0.4392 mol BrF3
3. Compare the number of moles of BrF3 needed with the actual number of moles of BrF3 given:
Actual number of moles of BrF3 = 0.4237 mol BrF3
If the actual number of moles of BrF3 given is less than the required number of moles (0.4392 mol), then BrF3 is the limiting reactant and I2O5 is in excess.
To find the mass of the excess reagent (I2O5), we can use the molar mass of I2O5:
Molar mass of I2O5 = 2 (126.9 g/mol) + 5 (16.0 g/mol) = 333.8 g/mol
Mass of excess reagent (I2O5) = (0.1318 mol I2O5) * (333.8 g/mol) = 43.99 g
Therefore, the mass of the excess reagent is approximately 43.99 grams.