Answer to Question #312937 in General Chemistry for Ndiza

Question

Consider the reaction below

I2O5(g) + 5 CO2(g) ———> 5 CO2(g) + I2(g)

80.0g of iodine (v) oxide reacts with 28.0g of carbon monoxide. Determine the mass of iodine I2 which would be produced

Answer 1

Solution:

The molar mass of iodine(V) oxide (I₂O₅) is 333.81 g/mol

The molar mass of carbon monoxide (CO) is 28.01 g/mol

Calculate moles of each reactant:

Moles of I₂O₅ = (80.0 g I₂O₅) × (1 mol I₂O₅ / 333.81 g I₂O₅) = 0.23966 mol I₂O₅

Moles of CO = (28.0 g CO) × (1 mol CO / 28.01 g CO) = 0.99964 mol CO

Balanced chemical equation:

I₂O₅(g) + 5CO(g) ⟶ 5CO₂(g) + I₂(g)

According to stoichiometry:

1 mol of I₂O₅ reacts with 5 mol of CO

Thus, 0.23966 mol I₂O₅ reacts with:

(0.23966 mol I₂O₅) × (5 mol CO / 1 mol I₂O₅) = 1.1983 mol CO

However, initially there is 0.99964 mol of CO (according to the task).

Therefore, CO acts as limiting reactant and I₂O₅ is excess reactant.

According to stoichiometry:

5 mol of CO produces 1 mol of I₂

Thus, 0.99964 mol of CO produces:

( 0.99964 mol CO) × (1 mol I₂ / 5 mol CO) = 0.19993 mol I₂

The molar mass of iodine (I₂) is 253.809 g/mol

Therefore,

Mass of I₂ = (0.19993 mol I₂) × (253.809 g I₂ / 1 mol I₂) = 50.744 g I₂ = 50.7 g I₂

Mass of I₂ = 50.7 g

Answer: 50.7 grams of iodine (I₂) would be produced