Question

Calculate the theoretical yield in , grams AlI₃, aluminum iodide, from the complete reaction of 113 grams I₂ according to the following balanced chemical equation: 2Al(s) + 3I₂(s) → 2AlI₃(s)

Answer 1

121 g $AlI_{3}$ aluminum iodide, from the complete reaction of 113 grams I₂ according to the following balanced chemical equation:

2Al(s) + 3I₂(s) → 2$AlI_{3}$

As per the given balanced equation,

3 moles of $I_{2}$ are used to give 2 moles of $AlI_{3}$.

Atomic weight of iodine is 127 gm/ mole

So, molecular weight of iodine molecule = 254 gm/ mole

There for , we can say that 1 mole o iodine molecule contain 254 g

So ,Given 113 gm iodine contains how many moles?

  Moles of iodine = 113/254 = 0.44 moles

from the equation we can say that 3 moles of iodine used to give 2 moles of aluminum iodide.

So, 0.44 moles of iodine gives how many moles of aluminum iodide ?

   Moles of aluminum iodide = 0.44 × 2 ÷  3

                                             = 0.2965 moles

Molecular weight of one mole of aluminum iodide = atomic weight of Al + 3( atomic weight of iodine )

      = 408 gm

∴ 1 mole of aluminum iodide = 408 gm

so, 0.2965 moles of aluminum iodide = ?

            = 0.2964 × 408

            = 121 gm  Aluminum iodide formed.

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