# Stoichiometry

### Stoich for Dummies

## Stoich Beginnings

- Write out the reaction in a chemical equation form. Do this by determining the reactants and the product depending on the type of reaction.
- After setting up the equation, you must balance it so there is an equal amount of elements on each side.
- Then find the molar mass for each reactant and product in the equation. MOLAR MASS means finding one molecule of said element or compound.

## Follow Along Problem

Reaction of Nitrogen and Hydrogen.

## First Step By writing out the reaction, you'll be able to see it easier. Just for reference, the chemical equation reads, Nitrogen + Hydrogen yields Ammonia. | ## Second Step Before we balance, find the molar mass of each reactant and product so that the coefficients don't confuse us during our calculations. | ## Third Step After finding the mass, the equation needs to be balanced so that you can obtain the necessary coefficients for dimensional analysis. |

## First Step

By writing out the reaction, you'll be able to see it easier.

Just for reference, the chemical equation reads, Nitrogen + Hydrogen yields Ammonia.

Just for reference, the chemical equation reads, Nitrogen + Hydrogen yields Ammonia.

## Second Step

Before we balance, find the molar mass of each reactant and product so that the coefficients don't confuse us during our calculations.

## Mole to Mole

- Use dimensional analysis to convert the given moles of an element into the wanted moles.
- Follow the dimensional analysis chart.

## Mole To Mole Diagram This diagram will help you with mole to mole conversions. | ## Example Problem From the previous balanced equation we were given 7.27 moles of nitrogen. The problem wants us to find the amount of moles of NH3 that can be made. |

## Mass To Mass Conversions

- Plug the given values into the diagram and solve by multiplying across then dividing.
- Make sure inputed mass is molar mass of one molecule.

## Mass to Mass Chart By following this chart, you'll be able to ace mass to mass conversions. | ## First Step In this problem the given was 12.1 grams of nitrogen which will continue to be Mole A throughout this example. The problem wants us to find how many grams of Mole B, which is our product, will be made by the given. | ## Last Step Check your significant figures and REMEMBER to put units on your answer. |

## First Step

In this problem the given was 12.1 grams of nitrogen which will continue to be Mole A throughout this example. The problem wants us to find how many grams of Mole B, which is our product, will be made by the given.

## Limiting and Excess Reactants

- Find the mass to mass conversion with both given amount of reactants.
- Compare the answers and whichever has the smaller amount is the limiting reactant while the other is the excess reactant

## Follow Mass to Mass Conversions Chart You'll have to follow this but instead of only doing it once, you'll have to do it twice | ## First Step The question will usually give you the given amounts to calculate the excess and limiting reactants. In this problem Mole A (Nitrogen) will have 12.1 grams and Mole A2 (Hydrogen, the second reactant) will have 20.15 grams. | ## Second Step Since we already calculated Mole A with 12.1 grams we can move on to Mole A2 with 20.15 grams. Try to follow the chart and use Mole A2 (Hydrogen) to find Mole B (Ammonia). Check your answer with the picture |

## Follow Mass to Mass Conversions Chart

You'll have to follow this but instead of only doing it once, you'll have to do it twice

## First Step

The question will usually give you the given amounts to calculate the excess and limiting reactants. In this problem Mole A (Nitrogen) will have 12.1 grams and Mole A2 (Hydrogen, the second reactant) will have 20.15 grams.

## Theoretical Yield and Percent Yield

Theoretical yield is just the yield of the product using the limiting reactant. So in this case the theoretical yield would be 17.7 grams of Ammonia

## Ammonia in the Real World

Ammonia is used in metallurgical processes because it can be decomposed easily to yield hydrogen for wielding. It can also absorb a high degree of heat therefore it's useful as a coolant.