Final answer:
The minimum volume of oxygen required for the complete combustion of 100 cm3 of propene is 450 cm3. The volume of carbon dioxide produced in this reaction is 300 cm3. The mass of water formed from the combustion of 100 cm3 of propene is 2.21 g.
Step-by-step explanation:
The student has asked about the combustion of propene (C3H6), but provided general information on combustion reactions with different hydrocarbons such as propane (C3H8). As the question pertains specifically to propene (C3H6), let's focus on its combustion reaction. The balanced chemical equation for the complete combustion of 2 moles of propene is:
2 C3H6 (g) + 9 O2 (g) -> 6 CO2 (g) + 6 H2O (l)
Using this balanced equation:
- To calculate the minimum volume of oxygen required for the complete combustion of 100 cm3 of propene, we use the molar ratio from the balanced equation where 2 volumes of propene will require 9 volumes of oxygen. Hence, for 100 cm3 of propene, the volume of oxygen required is (100 cm3 * 9/2) = 450 cm3.
- To calculate the volume of carbon dioxide produced, we use the ratio from the balanced equation, where 2 volumes of propene produce 6 volumes of CO2. So, for 100 cm3 of propene, the volume of CO2 produced is (100 cm3 * 6/2) = 300 cm3.
- For the mass of water formed, we have to use the molar mass of water (H2O), which is 18.01528 g/mol. From the balanced equation, 2 moles of propene yield 6 moles of water. Assuming all volumes are at the same temperature and pressure, 100 cm3 of propene (which is approximately 1/24.5 of a mole at RTP) would produce 3/24.5 moles of water. The mass of water produced would be (3/24.5 moles * 18.01528 g/mol) = 2.21 g.