C3H8 + 5O2 --> 3CO2 + 4H2O

Question

1. If 7.6 mol CO2 are produced, how many mole of oxygen re needed to carry out the reaction?
2. If 13.4 g H2O are produced, how many grams of propan are required to carry out the reaction?
3. If propane reacts with 31.2L O2, how many liters of carbon dioxide will be produced?
4. If 8298 grams of propane react with oxygen, how many liters of water vapor will be produced

Thank you to anyone that solves these!!! I appreciate you very much!!

Answer 1

1)

7.6 mol CO2 x (5 mol O2 / 3 mol CO2) = 12.7 mol O2

2)

13.4 g H2O x (1 mol C3H8 / 4 mol H2O) = 3.4 mol C3H8

3.4 mol C3H8 x (44.1g / mol) = 150g C3H8

3)

31.2L O2 / 22.4 = 1.39 mol O2 x (3 mol CO2 / 5 mol O2) = 0.83 mol CO2

0.83 mol CO2 x 22.4 = 18.6L CO2

4)

8298g C3H8 x (mol / 44.1g) = 188 mol C3H8 x (4 mol H2O / 1 mol C3H8) = 752 mol H2O

752 mol H2O x 22.4 = 16845L H2O

**The last answer (4) seems very high to me, but I am fairly confident I have done it correctly, so I would both double check the date from the problem and try to do this one yourself as well. But I hope this helps a little bit. Let me know if you need a further explanation for how to do these.