Question

The Lamborghini Aventador engine has a 12 -cylinder engine in which each cylinder has a volume of \(542 \mathrm{~cm}^{3} .\) Each cylinder is full of air at \(85^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\) (a) How many moles of oxygen are in each cylinder? \(\left(\right.\) mole percent of \(\mathrm{O}_{2}\) in air \(\left.=21.0\right)\) (b) Assume that the hydrocarbons in gasoline have an average molar mass of \(1.0 \times 10^{2} \mathrm{~g} / \mathrm{mol}\) and react with oxygen in a 1:12 mole ratio. How many grams of gasoline should be injected into the cylinder to react with the Oxygen?

Short answer

Answer: There are 0.004263 moles of oxygen in each cylinder, and 0.0355 grams of gasoline should be injected into the cylinder for complete combustion.

Step-by-step solution

Convert Celsius to Kelvin

Convert the given temperature from Celsius to Kelvin by adding 273 to the temperature: \(T = 85°C + 273 = 358K\).

Convert Volume to Liters

The given volume is in cm³, to convert it to liters, divide the volume by 1000: \(V_{L} = 542 \ cm^3 / 1000 = 0.542L\).

Calculate Moles of Air

Use the ideal gas law to calculate the moles of air in the cylinder: \(PV = nRT → n = PV/RT\) \(n = (1.00 \ atm)(0.542 \ L) / (0.0821 \ L.atm/(mol.K))(358 \ K)\) \(n = 0.0203 \ mol\)

Calculate Moles of Oxygen

Now, calculate the moles of Oxygen in the cylinder using the given mole percent of Oxygen in the air: Moles of O₂ = 21.0% of moles of air. Moles of O₂ = 0.210 * 0.0203 mol Moles of O₂ = 0.004263 mol #a) Answer: There are 0.004263 moles of oxygen in each cylinder.# #b) Calculate the grams of gasoline needed for complete combustion#

Calculate Moles of Gasoline

Use the given stoichiometric ratio of 1:12 mole ratio for gasoline and oxygen to find the moles of gasoline required: Moles of Gasoline = Moles of Oxygen / 12 Moles of Gasoline = 0.004263 mol / 12 Moles of Gasoline = 0.000355 mol

Calculate Grams of Gasoline

Now, use the given molar mass of gasoline (100g/mol) to calculate the mass of gasoline required for complete combustion: Mass of Gasoline = Moles of Gasoline * Molar Mass of Gasoline Mass of Gasoline = 0.000355 mol * 100g/mol Mass of Gasoline = 0.0355g #b) Answer: 0.0355 grams of gasoline should be injected into the cylinder to react with the oxygen.#