Al(s) + (g) → __ Allz (9) Using the unbalanced equation above, calculate the moles of aluminum iodide produced from 2.4 moles of iodine vapor,

Part A: Determine the limiting reagent and the theoretical yield of the product if one starts with:

1.20 mol Al and 2.40 mol iodine.

Solution: we already have moles as the unit, so we use those numbers directly.

Here is how to find out the limiting reagent…

Take the moles of each substance and divide it by the coefficient of the balanced equation.

For aluminum: 1.20 ÷ 2 = 0.60  The lowest number of moles indicates the limiting reagent.

For iodine: 2.40 ÷ 3 = 0.80 Aluminum will run out first in part A.

Theoretical yield of AlI3 (in moles) for part A is 1.20. Note the 1:1 mole ratio.

Part B: Determine the limiting reagent and the theoretical yield of the product if one starts with:

1.20 g Al and 2.40 g iodine.

2 Al + 3 I2 > 2 AlI3

Solution: since we have grams, we must first convert to moles. Then we solve just as we did in

part a just above (divide by the coefficient)

For the mole calculation:

aluminum is 1.20 g ÷ 26.98 g mol¯1 = 0.04477 mol

iodine is 2.4 g ÷ 253.8 g mol¯1 = 0.009456 mol

To determine the limiting reagent:

aluminum is 0.04477 ÷ 2 = 0.02238

iodine is 0.009456 ÷ 3 = 0.003152  The lower number of moles is iodine.

Finally, we have to do a calculation and it will involve the iodine, NOT the aluminum. Why?

From here figure out the grams of AlI3 and you have your answer.

0.009456 mol I2 x 2 AlI3/3I2 = 0.006304 mol AlI3 x 407.7g/mol AlI3 = 2.57 g

Theoretical yield of AlI3 is 2.57 g.

Explanation:

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