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Zinc reacts with iodine in a synthesis reaction. Using a balanced chemical equation for the reaction, determine the percent yield of a 125.0gram sample of zinc was used and 515.6 grams of product is recovered

Respuesta :

Zn + I2 = ZnI2


Percentage yield = actual yield / theoretical yield * 100


Actual yield of ZnI2 = 515.6g


Mole of zinc reacted = 125 / 65.4 = 1.91 mol


for every 1 mole of zinc will form 1 mole of ZnI2


Mole of ZnI2 formed = 1.91 * 1 = 1.91 mol


Theoretical yield of ZnI2 = 1.91 * (65.4 + 126.9*2) = 609.672 g


515.6 / 609.672 * 100 = 84.6%

dsdrajlin

Answer:

84.48%

Explanation:

Let's consider the following synthesis reaction.

Zn + I₂ → ZnI₂

We can establish the following relations:

  • The molar mass of Zn is 65.38 g/mol.
  • The molar ratio of Zn to ZnI₂ is 1:1.
  • The molar mass of ZnI₂ is 319.22 g/mol.

The mass of ZnI₂ (theoretical yield) to be obtained from 125.0 g of Zn is:

[tex]125.0gZn.\frac{1molZn}{65.38gZn} .\frac{1molZnI_{2}}{1molZn} .\frac{319.22gZnI_{2}}{1molZnI_{2}} =610.3gZnI_{2}[/tex]

The real yield of ZnI₂ is 515.6 g. The percent yield is:

(515.6 g/610.3 g) × 100% = 84.48%

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