Ozone reacts with nitric oxide according to the equation NO(g) + O3(g) → NO2(g) + O2(g) 0.66 g NO(g) was mixed with 0.72 g O3(g) - HSC - SSCE Chemistry - Question 9 - 2004 - Paper 1

Next, to calculate the number of moles of O3, we use the molar mass of O3, which is approximately 48 g/mol:

$n_{O3} = \frac{0.72 \ g}{48 \ g/mol} = 0.015 \ mol$

The balanced reaction shows a 1:1 molar ratio between NO and O3. Since we have 0.022 mol NO and only 0.015 mol O3, O3 is the limiting reactant.

From the stoichiometry of the reaction, 1 mol of O3 produces 1 mol of NO2. Therefore, the moles of NO2 produced is equal to the moles of O3:

$n_{NO2} = 0.015 \ mol$

Using the ideal gas law formula:

$PV = nRT$

Rearranging for V gives:

$V = \frac{nRT}{P}$

Where:

• P = 100 kPa = 100,000 Pa,
• R = 8.314 J/(mol·K) = 8.314 L·kPa/(mol·K),
• T = 0°C = 273.15 K.

Substituting the values:

$V = \frac{0.015 \ mol \times 8.314 \frac{L·kPa}{mol·K} \times 273.15 K }{100 \ kPa} \approx 0.34 \ L$