Question

Consider the reaction
I2O5(g) + 5 CO(g) -------> 5 CO2(g) + I2(g)
a) 80.0 grams of iodine(V) oxide, I2O5, reacts with 28.0 grams of carbon monoxide, CO.
Determine the mass of iodine I2, which could be produced?

Answer 1

Answer:

50.76 g of I2

Explanation:

Molecular mass of I2O5 = (126.9*2)+(16*5) g/mol = 333.8 g/mol

80.0 g of I2O5 = (80.0 g)/(333.8 g/mol) = 0.24 mol

Molecular mass of CO = (12)+(16) g/mol = 28 g/mol

28.0 g of CO = (28.0 g)/(28 g/mol) = 1 mol

1 mol of I2O5 requires 5 mol of CO

Hence 1 mol of CO requires 0.2 mol of I2O5    [CO is the limiting reagent]

1 mol of I2O5 produces 1 mol of I2

hence, 0.2 mol of I2 = (0.2 mol)(126.9*2 g/mol) = 50.76 g of I2