My questions is a general chemistry question that involves percent yields, theoretical yields, and the amounts of grams or moles of a substance required to process another substance.

Question

To monitor levels of ozone, O3, in the air, an air sample is processed by an instrument that causes ozone to react with sodium iodide in the following reaction: O3 (g) + 2NaI (aq) + H2O (l) yields O2 (g) + I2(s) + 2NaOH (aq)

a. Determine the number of moles of sodium iodide required to process 2.85 x 10 to the -6^th power moles of ozone. 

b. Determine how many grams of sodium iodide are required to process 3.00 grams of ozone. 

c. If 1455 g sodium iodide is available to process a sample of 250.0 g O3 in a reaction with excess water, determine the limiting reactant and the theoretical yield of iodine (I2). 

d. From question c, determine the number of iodine atoms present in the theoretical yield of iodine (I2). 

Answer 1

O₃ (g) + 2NaI (aq) + H2O (l) ==> O₂ (g) + I₂(s) + 2NaOH (aq) .. balanced equation

(a). 2.85x10^-6 mols O₃ x 2 mols NaI / mol O₃ = 5.70x10^-6 mols NaI needed

(b). 3 g O₃ x 1 mol / 48.0 g x 2 mol NaI / mol O₃ x 150 g NaI / mol = 18.8 g NaI needed

(c). 1455 g NaI x 1 mol NaI / 150 g = 9.7 mols NaI present

250.0 g O₃ x 1 mol O₃ / 48.0 g = 5.21 mols O₃ present

This makes NaI the limiting reactant as we need twice the number of mols NaI as O₃

Theoretical yield of I₂ = 9.7 mol NaI x 1 mol I₂ / 2 mol NaI x 254 g I₂ / mol = 1232 g I₂ = theoretical yld

(d). 1232 g I₂ x 1 mol / 254 g = 4.85 mol I₂ x 6.02x10²³ I₂ molecules/mol = 2.92x10²⁴ I₂ molecules

2.92x10²⁴ I₂ molecules x 2 I atoms / I₂ molecule = 5.84x10²⁴ I atoms