Iodine pentafluoride gas reacts with iodine fluoride gas producing iodine heptafluoride gas and iodine gas. What is the maximum number of grams of iodine gas that can be produced from a reaction of 10.0 g of iodine pentafluoride with 11.20 L of iodine fluoride gas at STP

63.45g of I₂ can be produced

Explanation:

IF₅ reacts with IF to produce IF₇ and I₂. The reaction is:

IF₅ + 2 IF → IF₇ + I₂

Moles of 10.0g of IF₅ (221.89g/mol):

10.0g IF₅ × (1mol / 221.89g) = 0.0451 moles of IF₅

Using PV / RT = n, it is possible to find moles of 11.20L of IF, thus:

1atm×11.20L / 0.082atmL/molK × 273K = 0.500 moles of IF.

At STP, pressure is 1atm, temperature is 273K and gas constant R is 0.082atmL/molK

For a complete reaction of IF₅ you need:

0.0451 moles of IF₅ × (2 moles IF / 1 mole IF₅) = 0.902 moles of IF. As you have just 0.500 moles of IF, the IF is the limiting reactant.

2 moles of IF produce 1 mole of I₂. 0.500 moles of IF produce:

0.500mol IF ₓ ( 1 mol I₂ / 2 mol IF) = 0.250 mol I₂

As molar mass of I₂ is 253.81g/mol, mass of 0.250 mol I₂ are:

0.250mol I₂ ₓ (253.81g / mol) =

gas and precipitate r the correct answers