Answer:
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%
