text: (a)  5.00 g of C3H8 / 44.1 g/mol = 0.113 moles\n\n0.113 * 5 = 0.565 moles\n\n10.0 g of O2 / 32.0 g/mol = 0.3125 moles, giving Oxygen as limiting reagent.\n\n(b)  1.00 g /44.1 g/mol = 0.0227 moles burned to give\n\n3 * 0.0227 moles = 0.0681 moles of CO2 formed\n\n0.0681 moles * 44.01 g/mole = 3.00 grams of CO2