Question

NaI + CI2 > NaCI + I2

A. What is my theoretical yield of I2 if I start with 543 g of NaI?

B. If I actually produced 117 g of I2, what is my percent yield?

Answer 1

A. mole ratio= 2:1=2:1

if 543g of NaI react= 543/300=1.81moles

Therefore, the number of moles of I2 produced is, 1.81/2 =0.905moles

0.905 moles=229.87g

B.$Percentage yeild =117/229.89*100$ =51.5%