The balanced chemical equation for the combustion of propane C3H8 is:C3H8 + 5O2 3CO2 + 4H2OTo find the volume of oxygen gas O2 required for complete combustion of 10 grams of propane, we first need to determine the number of moles of propane and then use stoichiometry to find the moles of oxygen gas required.The molar mass of propane C3H8 is: 3 12.01 g/mol for C + 8 1.01 g/mol for H = 44.09 g/molNow, we can find the moles of propane:moles of propane = mass of propane / molar mass of propanemoles of propane = 10 g / 44.09 g/mol = 0.2268 molesFrom the balanced chemical equation, we can see that 1 mole of propane requires 5 moles of oxygen gas for complete combustion. Therefore, we can find the moles of oxygen gas required:moles of O2 = moles of propane 5 moles O2 / 1 mole C3H8 moles of O2 = 0.2268 moles 5 = 1.134 molesNow, we can use the ideal gas law to find the volume of oxygen gas required at standard temperature and pressure STP . At STP, 1 mole of any gas occupies 22.4 liters. Therefore:volume of O2 = moles of O2 volume per mole at STPvolume of O2 = 1.134 moles 22.4 L/mol = 25.4 litersSo, the volume of oxygen gas O2 required for complete combustion of 10 grams of propane is approximately 25.4 liters.