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------------------------------------------------ Calculate the theoretical yield of [tex]ZnI_2[/tex] when 2.00 g of [tex]Zn[/tex] is added to 2.00 g of [tex]I_2[/tex].

What is the limiting reactant?

Answer :

Final answer:

The theoretical yield of ZnI2 is 7.76 g. The limiting reactant in this reaction is I2.

Explanation:

To calculate the theoretical yield of ZnI2, we need to determine the limiting reactant. First, we convert the masses of both Zn and I2 to moles using their molar masses. The molar mass of Zn is 65.38 g/mol and the molar mass of I2 is 253.8 g/mol. The number of moles of Zn is obtained by dividing its mass (2.00 g) by its molar mass, giving us 0.0306 mol. Similarly, the number of moles of I2 is 0.0079 mol. From the balanced equation Zn + I2 → ZnI2, we see that the ratio of Zn to ZnI2 is 1:1. Therefore, the number of moles of ZnI2 that can be produced is also 0.0306 mol. Multiplying by its molar mass, we find that the theoretical yield of ZnI2 is 7.76 g.

The limiting reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product that can be obtained. To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. In this case, the moles of Zn (0.0306 mol) is greater than the moles of I2 (0.0079 mol). Therefore, I2 is the limiting reactant.

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