Chemistry question

Question

80.0g of I₂O₅ reacts with 25.0g of CO. If 40.61g of I₂ are produced from the reaction, what is the % yield for the reaction.

I₂O₅ + 5CO → 5CO₂ + I₂

a)How many moles of iodine should have been produced?

b)What is your theoretical yield of iodine in grams? 

c)What is the % yield for the reaction? 

Answer 1

I₂O₅ + 5CO → 5CO₂ + I₂ ... balanced equation

First thing to do is... Find the limiting reactant:

80.0 g I₂O₅ x 1 mol I₂O₅/333.81 g = 0.2397 moles (÷1->0.2397)

25.0 g CO x 1 mol CO/28.01 g = 0.8925 moles (÷5->0.1785) ==> LIMITING REACTANT

a) 0.8925 moles CO x 1 mol I₂ / 5 mol CO = 0.179 moles I₂ should be produced

b) Theoretical yield in g = 0.179 moles I₂ x 253.8 g/mol = 45.4 g I₂

c) yield of reaction = actual/theoretical (x100%) = 40.61g/45.4 g (x100%) = 89.4%