# Stoichiometry

### Ex. Propane & Oxygen

## Balanced Equation & Molar Mass

**44.097**, O2 =

**31.998**, CO2 =

**44.009**, and

**18.015**.

## IUPAC Name

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.

Ex.

How much carbon dioxide can you make if you react 5.25 g of C3H8 with 12.18 g of O2?

Ans.

C3H8: 15.72 g of CO2 Excess Reactant

** O2: 10.05 g of CO2 Limiting Reactant**

Theoretical Yield = O2

## Percent Yield

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%**