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, we calculate the moles of O3 using its molar mass of approximately 48.00 g/mol:

$n_{O3} = \frac{0.72 \, g}{48.00 \, g/mol} \approx 0.015 \text{ moles O3}$

From the balanced equation, we see that 1 mole of NO reacts with 1 mole of O3. Comparing moles:

• Moles of NO = 0.022
• Moles of O3 = 0.015

Since there is less O3, it is the limiting reactant.

According to the balanced equation, 1 mole of O3 produces 1 mole of NO2. Thus, the moles of NO2 produced are equal to the moles of O3:

$n_{NO2} = n_{O3} = 0.015 \text{ moles}$

Using the ideal gas law, we can find the volume:

$PV = nRT$

Where:

• P = 100 kPa
• V = volume of NO2
• n = 0.015 moles
• R = 8.314 J/(mol·K) = 8.314 kPa·L/(mol·K)
• T = 273.15 K (0°C)

Rearranging for V gives us:

$V = \frac{nRT}{P} = \frac{0.015 \, mol \times 8.314 \, kPa \cdot L/(mol \cdot K) \times 273.15 K}{100 \, kPa}$

Calculating this yields:

$V \approx 0.34 \, L$