High School

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'll use the Ideal Gas Law and stoichiometry. Let's break it down into two parts:

Part A: Determine moles of [tex]O_2[/tex]


  1. Convert the temperature from Celsius to Kelvin:

    [tex]T(K) = 75 + 273.15 = 348.15 \, \text{K}[/tex]


  2. Use the Ideal Gas Law [tex]PV = nRT[/tex] to find the total moles of gas [tex]n[/tex]:


    • [tex]P = 99.3 \, \text{kPa} = 99.3 \, \text{kPa} = 99,300 \, \text{Pa}[/tex]

    • [tex]V = 525 \, \text{cm}^3 = 0.000525 \, \text{m}^3[/tex]

    • [tex]R = 8.314 \, \text{J/(mol·K)}[/tex]

    • [tex]T = 348.15 \, \text{K}[/tex]


    Rearrange the Ideal Gas Law to solve for [tex]n[/tex]:

    [tex]n = \frac{PV}{RT} = \frac{99,300 \, \text{Pa} \times 0.000525 \, \text{m}^3}{8.314 \, \text{J/(mol·K)} \times 348.15 \, \text{K}} \approx 0.0184 \, \text{mol}[/tex]


  3. Calculate the moles of [tex]O_2[/tex]:

    Since the mole fraction of [tex]O_2[/tex] in air is 0.209, the moles of [tex]O_2[/tex] is:

    [tex]n_{O_2} = 0.209 \times 0.0184 \, \text{mol} \approx 0.00385 \, \text{mol}[/tex]



Part B: Calculate the mass of [tex]C_8H_{18}[/tex] that can be combusted


  1. Balanced chemical equation for the combustion of octane ([tex]C_8H_{18}[/tex]):

    [tex]2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O[/tex]

    This tells us that 2 moles of [tex]C_8H_{18}[/tex] react with 25 moles of [tex]O_2[/tex].


  2. Determine moles of [tex]C_8H_{18}[/tex] that can be combusted:

    Using stoichiometry from the balanced equation:

    [tex]n_{C_8H_{18}} = \frac{2}{25} \times n_{O_2}[/tex]

    [tex]n_{C_8H_{18}} = \frac{2}{25} \times 0.00385 \, \text{mol} \approx 0.000308 \, \text{mol}[/tex]


  3. Calculate the mass of [tex]C_8H_{18}[/tex]:

    The molar mass of [tex]C_8H_{18}[/tex] is [tex](8 \times 12.01) + (18 \times 1.008) = 114.22 \, \text{g/mol}[/tex].

    [tex]m = n \times \text{molar mass} = 0.000308 \, \text{mol} \times 114.22 \, \text{g/mol} \approx 0.0352 \, \text{g}[/tex]



Therefore, about 0.0352 grams of [tex]C_8H_{18}[/tex] could be combusted by the given quantity of [tex]O_2[/tex], assuming complete combustion.

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