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
1 answer
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.)
2 answers