The maximum amount of IF7 that can be obtained from 25.0 g is fluorine is ________. I₂ + F₂ → IF7

Question

Quizplus · Accepted Answer

To find the maximum amount of IF7 that can be obtained, we need to determine the limiting reagent. We have 25.0 g of fluorine (F₂) but we don't know the amount of iodine (I₂) we have.

We can calculate the theoretical yield of IF7 using both the amounts of iodine and fluorine:

1 mole of I2 reacts with 1 mole of F2 to produce 2 moles of IF7.

First, we need to convert the mass of fluorine to moles:

25.0 g F2 x (1 mol F2/38.0 g F2) = 0.658 moles F2

Next, we need to determine the amount of iodine required to react with all the fluorine:

1 mole I2 reacts with 1 mole F2, so we need 0.658 moles of I2.

Now we need to convert the moles of I2 to grams:

0.658 moles I2 x (253.81 g I2/1 mole I2) = 167.2 g I2

Since we don't have that much iodine (we only know we have less), the limiting reagent is I₂, and the maximum amount of IF7 that can be produced is determined by the amount of I₂ we have.

The maximum theoretical yield of IF7 is 0.658 moles (because we have 0.658 moles of F2), which corresponds to:

0.658 mol IF7 x (263.89 g IF7/1 mol IF7) = 173.7g IF7

But we know that we don't have enough I₂ to react with all the F₂. So, the actual yield will be less.

The amount of I₂ needed to react with all the F₂ is:

0.658 moles F2 x (1 mole I2 / 1 mole F2) = 0.658 moles I2

This corresponds to:

0.658 mol I2 x (253.81 g I2/1 mol I2) = 167.2 g I2

So, the maximum amount of IF7 that can be obtained is limited by the amount of iodine we have, which is less than 167.2 g.

Therefore, the best answer is B) 48.9 g.

The Maximum Amount of IF7 That Can Be Obtained from 25.0