Question

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

Answer 1

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%

Answer 2

84.48%

Step-by-step 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:

`125.0gZn.(1molZn)/(65.38gZn) .(1molZnI_(2))/(1molZn) .(319.22gZnI_(2))/(1molZnI_(2)) =610.3gZnI_(2)`

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

(515.6 g/610.3 g) × 100% = 84.48%