Question

Balance the following reaction.  

If we start with 34.1 g of diiodine pentoxide and react with 20.0 g of carbon monoxide, what is the theoretical yield of iodine and what is the limiting reagent?  

If 20.1 g of iodine is actually obtained, what is the % yield?  

What volume would the 20.0 g of carbon monoxide occupy at standard temperature and pressure?  carbon monoxide + diiodine pentoxide → iodine + carbon dioxide

Answer 1

a) 5CO + I2O5 -----> I2 + 5 CO2

b) 5 moles of CO(5x28g/mol) reacts with one mole of I2O5 (334g/mol)

Thus to react with 20 g of CO we need 46.76 g of I2O5

But we have 34.1 g of I2O5 only. thus

I2O5 is the limiting reagent.

Thus the product is calculated using limiting reagent only.

thus one mole of I2 O5 (334g/mol) gives one mole of I2 (254g/mol)

34.1 g of I2O5 gives = 25.93 g of Iodine

The theoretical yield of iodine is 25.93g

c) experimental yield = 20.1 g

Thus % yield = experimental yield x 100 / theoretical yield

= 20.1 x100 / 25.93

=77.52 %

d) PV = nRT for an ideal gas

using the same forCO, and substituting P = 1 atm, n = 20/28 , T = 273 K and R = 0.0821

Volume = nRT / P = (20/28) 0.0821 x 273/1

= 16.0095 L