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Assume that a single cylinder of an automobile engine has a volume of 525 cm³.

Part A
If the cylinder is full of air at 75°C and 99.3 kPa, how many moles of O₂ are present? (The mole fraction of O₂ in dry air is 0.209)
n = ? mol

Part B
How many grams of C₈H₁₈ could be combusted by this quantity of O₂, assuming complete combustion with formation of CO₂ and m = ? g

Answer :

To solve this problem, we need to use the ideal gas law and basic stoichiometry. Let's break it down step by step.

Part A: Calculate the moles of O₂ present in the cylinder.

  1. Convert known values to appropriate units:

    • Volume (V) = 525 cm³ = 0.525 L (since 1 L = 1000 cm³)
    • Temperature (T) = 75°C = 348 K (since K = °C + 273)
    • Pressure (P) = 99.3 kPa = 99.3 kPa / 101.3 kPa = 0.980 atm (since 1 atm = 101.3 kPa)

  2. Use the Ideal Gas Law:

    The formula is:

    [tex]PV = nRT[/tex]

    Where:
    [tex]P[/tex] is the pressure in atm, [tex]V[/tex] is the volume in liters, [tex]n[/tex] is the number of moles, [tex]R[/tex] is the ideal gas constant (0.0821 L·atm/mol·K), and [tex]T[/tex] is the temperature in Kelvin.

    First, we find the total moles of air in the cylinder using the whole cylinder volume:

    [tex]n_{\text{total}} = \frac{PV}{RT} = \frac{(0.980 \, \text{atm})(0.525 \, \text{L})}{(0.0821 \, \text{L·atm/mol·K})(348 \, \text{K})}[/tex]

    Solving this gives:

    [tex]n_{\text{total}} \approx 0.018 \text{ mol}[/tex]

  3. Calculate moles of O₂:

    Since the mole fraction of O₂ in air is 0.209, we multiply the total moles by this fraction:

    [tex]n_{\text{O}_2} = 0.018 \, \text{mol} \times 0.209 \approx 0.003762 \, \text{mol}[/tex]

Part B: Calculate the grams of C₈H₁₈ that could be combusted.

  1. Determine the stoichiometry of the reaction:

    The balanced chemical equation for the combustion of octane (C₈H₁₈) is:

    [tex]2\text{ C}_8\text{H}_{18} + 25\text{ O}_2 \rightarrow 16\text{ CO}_2 + 18\text{ H}_2\text{O}[/tex]

    From the equation, 2 moles of C₈H₁₈ require 25 moles of O₂ to combust completely.

  2. Calculate how many moles of C₈H₁₈ could be combusted:

    Using the stoichiometric ratio from the balanced equation:

    [tex]\frac{2 \text{ mol C}_8\text{H}_{18}}{25 \text{ mol O}_2} \times 0.003762 \text{ mol O}_2 = 0.000301 \text{ mol C}_8\text{H}_{18}[/tex]

  3. Convert moles of C₈H₁₈ to grams:

    The molar mass of C₈H₁₈ is approximately 114.22 g/mol (C: 12.01 g/mol * 8 + H: 1.008 g/mol * 18).

    [tex]0.000301 \text{ mol C}_8\text{H}_{18} \times 114.22 \text{ g/mol} \approx 0.0344 \text{ g}[/tex]

So, approximately 0.0344 grams of C₈H₁₈ can be combusted with the available O₂ in the cylinder.

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