Question

5.1 Ozone (03) reacts with NO to produce NO₂ gas. NO forms largely as a result of emissions from exhaust gases. The following reaction takes place 0₂ + NO2(g) 03(g) + NO(g) → In one such reaction, 0.74g 03 reacts with 0.67g NO a. Calculate the number of moles of 03 and of NO present at the start of the reaction b. Identify the limiting reactant c. Calculate the mass of NO2 produced​

Answer 1

To calculate the number of moles of O3 and NO at the start of the reaction:

a. Calculate the number of moles of O3:

Number of moles = mass of O3 (g) / molar mass of O3 (g/mol)

The molar mass of O3 (ozone) is approximately 48 g/mol.

Number of moles of O3 = 0.74g / 48 g/mol

b. Calculate the number of moles of NO:

Number of moles = mass of NO (g) / molar mass of NO (g/mol)

The molar mass of NO is approximately 30 g/mol.

Number of moles of NO = 0.67g / 30 g/mol

c. To identify the limiting reactant, compare the moles of O3 and NO. The reactant that produces fewer moles of product (NO2) is the limiting reactant.

d. Now, let's calculate the mass of NO2 produced using stoichiometry. The balanced equation is:

2 O3(g) + 2 NO(g) → 2 NO2(g) + O2(g)

First, find the limiting reactant (the one that produced fewer moles). Once you have the limiting reactant, use the stoichiometric ratios to find the moles of NO2 produced.

After calculating the moles of NO2 produced, multiply it by the molar mass of NO2 (approximately 46 g/mol) to get the mass of NO2 produced.