The combustion of 75.5 liters of 2- propane (c3h6)with 95.5 liters of oxygen gas is extremely exothermic. What is the maximum amount of water (in grams) to be made from this and how much of the non-limiting reactant is unused at the end (in liters)?

Answer 1

The equation is (note no lower case letters used)
2C3H6 + 9O2 ==> 6CO2 + 6H2O

Convert L C3H6 to L CO2.
75.5L x (6 mols CO2/2 moles C3H6) = 226.5 L CO2.

Convert 95.5 L O2 to L CO2.
95.5 x (6 moles CO2/9 moles O2) = 63.7 L CO2.
In limiting reagent problems, the smaller value of the product formed is ALWAYS the correct value; therefore, O2 is the limiting reagent.
This should serve as an example; you can do it for H2O. Use the same kind of process to determine how much of the C3H6 was used, then subtract from the initial amount to determine the amount not reacted.