Assume that a single cylinder of an automobile engine has a volume of $524 \mathrm{~cm}^{3}\(. (a) If the cylinder is full of air at \)74^{\circ} \mathrm{C}\( and \)99.3 \mathrm{kPa}\(, how many moles of \)\mathrm{O}_{2}$ are present? (The mole fraction of \(\mathrm{O}_{2}\) in dry air is \(0.2095 .\) ) (b) How many grams of \(\mathrm{C}_{8} \mathrm{H}_{18}\) could be combusted by this quantity of \(\mathrm{O}_{2}\), assuming complete combustion with formation of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?\)

We will use the Ideal Gas Law to find the total number of moles of air in the cylinder: \(PV = nRT\) Where P is the pressure (in atm), V is the volume (in L), n is the number of moles, R is the gas constant (0.0821 atm L/mol K), and T is the temperature (in Kelvin). First, we need to convert the given values: - Volume (1 cm³ = 0.001 L): 524 cm³ = 0.524 L - Temperature (Kelvin = Celsius + 273.15): 74°C = 347.15 K - Pressure (1 kPa = 0.009869 atm): 99.3 kPa = 0.98644 atm Now, rearrange the ideal gas law equation to solve for n: \(n = \frac{PV}{RT}\) Plug in the given values and calculate n: \(n = \frac{(0.98644)(0.524)}{(0.0821)(347.15)} = 0.0209 \, \mathrm{moles \, of \, air}\)

Now, we will use the mole fraction of O₂ in the air to find the number of moles of O₂ in the cylinder: \[n_{O_2} = n_{\mathrm{air}} \times X_{O_2}\] Where nₐᵢᵣ is the number of moles of air and X_O₂ is the mole fraction of O₂. Plug in the values and calculate the number of moles of O₂: \(n_{O_2}= (0.0209)(0.2095)= 0.00438\, \mathrm{moles\, of\, O_2}\) Now, we have the number of moles of O₂ in the cylinder: \(0.00438 \, \text{moles}\).

The balanced combustion reaction for C₈H₁₈ with O₂, forming CO₂ and H₂O is: \[ C_8H_{18} + 12.5O_2 \to 8CO_2 + 9H_2O\] From the balanced equation, we can see that 12.5 moles of O₂ react with 1 mole of C₈H₁₈. Now, set up the ratio to find the number of moles of C₈H₁₈: \( \frac{n_{C_8H_{18}}}{n_{O_2}}=\frac{1}{12.5}\) Solve for n_C8H18: \(n_{C_8H_{18}} = \frac{1}{12.5} \times 0.00438=\ 0.000350\, \mathrm{moles\, of\, C_8H_{18}}\) Finally, calculate the mass of C₈H₁₈ using its molar mass (C=12.01 g/mol, H=1.01 g/mol): \( M_{C_8H_{18}}=12.01 \times 8 + 1.01 \times 18 = 114.22\, \mathrm{g/mol}\) Now, calculate the grams of C₈H₁₈: \(g_{C_8H_{18}}=0.000350\,\mathrm{moles\, of\, C_8H_{18}} \times 114.22\, \mathrm{g/mol}\) \(g_{C_8H_{18}} \, \approx \, 0.04 \, \mathrm{g}\) So, about 0.04 grams of C₈H₁₈ can be combusted by the given quantity of O₂ in the cylinder, assuming complete combustion with the formation of CO₂ and H₂O.