In this reaction: Mg (s) + I₂ (s) → MgI₂ (s) If 3.34 moles of Mg react with 3.56 moles of I₂, and 1.76 moles of MgI₂ form, what is the percent yield?

52.7%

Explanation:

The balanced equation for the reaction is given below:

Mg (s) + I₂ (s) → MgI₂ (s)

From the balanced equation above,

1 mole of Mg reacted with 1 mole of I₂ to produce 1 mole of MgI₂.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted with 1 mole of I₂.

Therefore, 3.34 moles of Mg will also react with 3.34 moles of I₂.

From the illustration made above, we can see that only 3.34 moles out of 3.56 moles of I₂ reacted completely with 3.34 moles of Mg.

Therefore, Mg is the limiting reactant and I₂ is the excess reactant.

Next, we shall determine the theoretical yield of MgI₂.

In this case, the limiting reactant will be use because it will produce the maximum yield of MgI₂ as all of it is consumed in the reaction.

The limiting reactant is Mg and the theoretical yield of MgI₂ can be obtained as follow:

From the balanced equation above,

1 mole of Mg reacted to produce 1 mole of MgI₂.

Therefore, 3.34 moles of Mg will also react to produce 3.34 moles of MgI₂.

Thus, the theoretical yield of MgI₂ is 3.34 moles.

Finally, we shall determine the percentage yield of MgI₂ as follow:

Actual yield of MgI₂ = 1.76 moles

Theoretical yield of MgI₂ = 3.34 moles.

Percentage yield of MgI₂ =.?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 1.76/3.34 × 100

Percentage yield of MgI₂ = 52.7%

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