Question

In this reaction: Mg (s) + 12 (s) → Mg 2 (s), if 10.0 g of Mg reacts with 60.0 g of 12, and 61.26 g of Mgl2 form, what is the percent yield?

Answer 1

Number of moles of Mg = 10.0 g / 24.305 g/mol = 0.411 mole

Number of moles of I2 = 60.0 g / 253.8089 g/mol = 0.236 mole

From the balanced equation we can say that

1 mole of Mg requires 1 mole of I2 so

0.411 mole of Mg will require

= 0.411 mole of Mg *(1 mole of I2 / 1 mole of Mg)

= 0.411 mole of I2

But we have 0.236 mole of I2 which is in short so I2 is limiting reactant

From the balanced equation we can say that

1 mole of I2 produces 1 mole of MgI2 so

0.236 mole of I2 will produce

= 0.236 mole of I2 *(1 mole of MgI2 / 1 mole of I2)

= 0.236 mole of MgI2

mass of 1 mole of MgI2 = 278.1139 g so

mass of 0.236 mole of MgI2 = 65.6 g

Therefore, theoretical yield of MgI2 = 65.6 g

percent yield = (actual yield / theoretical yield)*100

percent yield = (61.26 / 65.6)*100

percent yield = 93.4 %