Final answer:
To find the volume of H2O produced from the combustion of 50.0 g of octane, we first balance the chemical equation, convert grams of octane to moles, use the stoichiometry to find moles of water, and apply the ideal gas law, resulting in approximately 88.2 liters of H2O vapor.
Step-by-step explanation:
The question is asking for the volume of water vapor produced from the combustion of octane (C8H18), which is a chemistry problem involving stoichiometry and the use of the ideal gas law. To solve this, we first need a balanced chemical equation for the complete combustion of octane, which is C8H18 + 12.5O2 → 8CO2 + 9H2O. Next, we use the molar mass of octane (114.23 g/mol) to convert 50.0 g of octane to moles. Then, we use the stoichiometry of the reaction to find the moles of water produced. Finally, we apply the ideal gas law, assuming standard conditions (0°C and 1 atm), to convert the moles of water to liters. Remember under these conditions, 1 mole of any ideal gas occupies 22.4 liters.
First, convert 50.0 g octane to moles:
50.0 g ÷ 114.23 g/mol = 0.43757 mol C8H18. Then apply the stoichiometry:
9 mol H2O / 1 mol C8H18 × 0.43757 mol C8H18 = 3.93813 mol H2O. Now, convert moles of water to liters at standard conditions:
3.93813 mol × 22.4 L/mol = 88.17412 L of H2O vapor. The volume of gaseous H2O produced from burning 50.0 g of octane is closest to 88.2 liters.