I2O5+CO>> I2 + CO2
balance...
I2O5 + 5CO>>I2 + 5CO2
so the next is to see how many moles you have
28gCO/28= 1mol CO
80/334= .24 moles.
Since you need 5 mole CO per mole of I2O5, then I2O5 is limiting, and you have CO in excess.
so you will get .24mol I2, convert that to grams.
80.0 g of iodine (v) oxide reacts with 28.0 g of carbon monoxide determine the mass of iodine
1 answer