1 C3H8 + 5 O2 --> 3 CO2 + 4H20. If 1.5 moles of C3H8 react, how many moles of CO2 will be produced? *
3.0 moles
4.5 moles
06 moles
O 8.0 moles

First confirm the reaction is balanced:

C3H8 + 5O2 --> 3CO2 + 4H20 (3 cabon - check; 8 hydrogen - check; 10 oxygen - check).

a) In the equation there is a 5:1 ratio between propane and oxygen.  We also know that number of mole is proportional to pressure and volume.  Since pressure is constant (STP) then the volume of O2 is 7.2 * 5 = 36 litres.

b) For a near ideal gas that PV = nRT (combined gas law).  So for 7.2 litres propane we find n(propane) = 101.3 * 7.2/8.314*298 ~ 0.29 mole (using metric units throughout for simplicity).

There is a 1:3 ratio between propane and CO2.  Therefore 3 * 0.29 = 0.87 mole of CO2 is produced.

MW(CO2) ~ 44 g/mol.  Therefore m(CO2) = 44 * 0.87 ~ 38.3 g

c) We know we need more oxygen than propane (due to the 1:5 ratio) so oxygen is the limiting reagent.  Again Volume is proportional to number of mole and we see there is a 5:4 ratio between oxygen and water.  Therefore the volume of water vapour produced will be (4/5) * 15 = 12 litres.

The other questions use the same technique and will give you some much needed practice.

Explanation: