Question

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

Answer 1

63.45g of I₂ can be produced

Step-by-step 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) =

63.45g of I₂ can be produced