# Stoich project

### James Chow 7th Period

## What is stoichiometry

## Example reaction and information

Balanced equation: 2KCl(aq) + 3O2(g) = 2KClO3(aq)

Reaction type: synthesis

IUPAC Names

KCl: Potassium chloride

O2: Oxygen

KClO3: Potassium chlorate

Molar Masses

Potassium chloride: 74.551 g/mol

Oxygen: 31.98 g/mol

Potassium chlorate: 122.548 g/mol

## Proof of reaction

## Mole to mole conversions

For example, we have 5.6 moles of KCl and we want to convert it to KClO3. Both of them have a coefficient of 2, so the ratio is 2 to 2. Therefore, KClO3 also has 5.6 moles.

## Mass to mass conversions

For example, you have 12.1 grams of KCl and you wish to convert it to KClO3. To solve this, you would divide 12.1 by 74.551, which is the molar mass. Once again the ratio is 2 to 2, so the number stays the same. You now finally multiply by 122.548, which is the molar mass of KClO3. Therefore, the answer is 19.89 grams of KClO3

## Limiting and excess reactant

For example, you have 12.3 grams of KCl and O2. After working out the mass to mass conversion, KCl can produce 20.219 grams of KClO3 while O2 can produce 31.404 grams of KClO3. Therefore, KCl is the limiting reactant because it can produce less, and O2 is the excess reactant because it can produce more than KCl can.

## Theoretical and percent yield

For example, the theoretical yield from the last question is 20.219 grams because you only have enough grams of KCl to produce that much. Even though O2 is capable of producing 31.404, KCl cant produce that much, so the answer will be the 21.219.

Now, lets say your actual yield is 17.359. 17.359 divided by 21.219 x 100 = 85.854. 85.854% is your percent yield.