Question

Q5) The reaction for the combustion of propane is as follows:

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

1. If 20.0 g of C3H8 and 20.0 g of O2 are reacted, which is the limiting reactant?
2. How many grams of H2O can be produced?
3. How many grams of the excess reactant remains at the end of the reaction?

Answer 1

Write the equation and balance it.
2a. Convert 20.0 g C3H8 to moles.
2b. Convert 20.0 g O2 to moles.

3a. Using the coefficients in the balanced equation, convert moles C3H8 to moles H2O.
3b. Same procedure convert moles O2 to moles H2O.
3c. You have two different answers and both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that number is the limiting reagent.
4. Using the smaller number, convert moles to grams. g = moles x molar mass.

To calculate grams excess reagent remaining, use step 3a or 3b above to convert moles of the non-limiting reagent to moles of limiting reagent, convert to grams, and subtract from 20.0.