Ex. Propane & Oxygen
Balanced Equation & Molar Mass
Ex. C3H8 - Propane, O2 - Oxygen, CO2 - Carbon Dioxide, H2O - Water
Type of Reaction
Mole to Mole Conversions
Ex 1. You have 11.30 moles of C3H8. How many moles of CO2 can you make?
Ans. (11.3 x 3) / 1 = 33.9 mole of CO2
Ex. 2 You want to make 12.1 moles of H2O. How many moles of O2 do you need?
Ans. (12.1 x 5) / 4 = 15.1 mole of O2
Mass to Mass Conversions
- Find the given. Which is the grams of mole A.
- Then find the molar mass of the mole A and divided it by mole A.
- Then find the ratio of mole B to mole A and multiply it to step 2.
- Then multiply the molar mass of mole B to step 3.
- And that is your answer.
Ex. How many grams of CO2 can you make when you use 8.12 grams of C3H8?
Ans. 8.12 grams of C3H8 / 44.097 mole of C3H8 = .184 grams per mole of C3H8 X (3 mole CO2 / 1 mole C3H8) = .552 X 44.009 mole of CO2 = 24.31 grams of CO2
Limiting & Excess Reactant & Theoretical Yield
- Find the two reactant that has mass with it.
- Set up the same equation with mass of each and put it in different equation.
- The lower number of the same amount of product is the limiting reactant.
- The higher one is excess reactant.
- To find the Theoretical Yield is the limiting reactant.
How much carbon dioxide can you make if you react 5.25 g of C3H8 with 12.18 g of O2?
C3H8: 15.72 g of CO2 Excess Reactant
O2: 10.05 g of CO2 Limiting Reactant
Theoretical Yield = O2
Ex. 8.4 g of CO2 were recovered from the experiment. What is your percent yield?
Ans. (8.4 / 10.1) X 100 = 83.17%