Gasoline (which can be considered to be octane, C8H18) burns in oxygen to produce carbon dioxide and water. What volume (L) of oxygen at STP is necessary to react with 1.0 gal of gasoline? (The density of gasoline is 0.81 g/mL. 1 gal = 3.78 L)

To determine the volume of oxygen gas required to react with 1.0 gallon of gasoline under standard temperature and pressure (STP) conditions, we need to use stoichiometry.

Step 1: Convert the volume of gasoline to mass. Density = Mass / Volume We have the volume (1.0 gal) and the density (0.81 g/mL), so we can calculate the mass: 1.0 gal = 3.78 L (given) 3.78 L * 1000 mL/L = 3780 mL (converting liters to milliliters) Density of gasoline = 0.81 g/mL Mass of gasoline = Density × Volume = 0.81 g/mL × 3780 mL = 3061.8 g

Step 2: Convert the mass of gasoline to moles using the molar mass of octane (C8H18). Octane has a molar mass of (8 × 12.01 g/mol for C) + (18 × 1.008 g/mol for H) ≈ 114.2 g/mol. Moles of octane = Mass / Molar mass = 3061.8 g / 114.2 g/mol ≈ 26.81 mol

Step 3: Write the balanced chemical equation for the combustion of octane: 2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(l)

Step 4: Use the stoichiometry of the balanced equation to find the moles of O2 needed. According to the balanced equation, 2 moles of octane (C8H18) require 25 moles of oxygen (O2) to react completely. Therefore, for 26.81 moles of octane, the moles of oxygen needed are: (25 moles O2 / 2 moles C8H18) × 26.81 moles C8H18 ≈ 335.13 moles O2

Step 5: Convert the moles of oxygen to volume using the molar volume of an ideal gas at STP, which is 22.414 L/mol. Volume of O2 = Moles × Molar volume = 335.13 moles × 22.414 L/mol ≈ 7517.32 L

Therefore, approximately 7517.32 liters of oxygen gas at STP are necessary to react with 1.0 gallon of gasoline.