# Learning Stoichiometry

### A step by step guide on solving stoichiometry problems

## Balancing Equations

Once you've written out your reactants and products, imagine there's a 1 coefficient in front of the largest or most complex product or reactant. Using the subscripts, balance the equation, making sure there's the same amount of each element on both side. Coefficients multiply with subscripts (i.e 3Ag2 is equal to 6Ag)

- Al + Ag2S yields Ag + Al2S3
**2**Al +**3**Ag2S yields**6**Ag +**1**Al2S3

## IUPAC name for each reactant and product

Use the chemical's name rather than it's symbol. Omit subscripts and coefficients.

- 2Al + 3Ag2S yields 6Ag + Al2S3
- Aluminum and silver sulfide yields silver and aluminum sulfide

## Finding Molar Mass

Using the molar mass of each element, calculate the molar mass of each product and reactant. Ignore coefficients, but multiply by the subscripts [i.e. Al=26.982 and 2Al=26.982 but Al2=2(26.982)]

- 2Al + 3Ag2S yields 6Ag + Al2S3
- 26.982g/mole + 215.736g/mole yields 107.868g/mole + 150.162g/mole

## Mole to Mole Conversions

Plug the given for mole A into the chart to solve for mole B. Multiply by the top, divide by the bottom:

Mole A Mole B(coefficient from equation)

-----------Mole A(coefficient from equation)

1. 3.28 mole Al 1 mole Al2S3 =

----------------------2 mole Al

Mole A Mole B(coefficient from equation)

-----------Mole A(coefficient from equation)

1. 3.28 mole Al 1 mole Al2S3 =

**1.64 mole Al2S3**----------------------2 mole Al

## Mass to Mass Conversions

Plug the given for mass A into the chart to solve for the mass B. Multiply by the top, divide by the bottom.

Mass A Mole A Mole B(coefficient from equation) Molar Mass B

------------Molar Mass A Mole A(coefficient from equation) Mole B

1. 12.1g Al 1 mole Al 3 mole Ag2S 215.736g Ag2S =

---------------- 26.982g Al 2 mole Al 1 mole Ag2S

**In the second and fourth column, the number of moles is always 1**:Mass A Mole A Mole B(coefficient from equation) Molar Mass B

------------Molar Mass A Mole A(coefficient from equation) Mole B

1. 12.1g Al 1 mole Al 3 mole Ag2S 215.736g Ag2S =

**145.12g Ag2S**---------------- 26.982g Al 2 mole Al 1 mole Ag2S

## Limiting and Excess Reactant (and Theoretical Yield)

The equation for solving is the same as mass to mass conversion, except two equations will be set up because there will be two givens. Once solved, the answer that's smaller is called the

**theoretical yield**. The given used in that equation is known as the**limiting reactant**because it ran out first, and the other given is known as the**excess reactant**because it was in excess or still had a remainder after the reaction.- Given:
**12.3g Al (Excess)**. Mass of Al2S3= 34.23g - Given:
**12.3g Ag2S (Limiting)**. Mass of Al2S3=**2.85g Al2S3 (Theoretical)**

## Percent Yield

Divide the actual yield by the theoretical yield, then multiply by 100 for percent yield

(actual yield/theoretical yield)100 = Percent Yield

(actual yield/theoretical yield)100 = Percent Yield

- (4.29/2.85)100 =
**150.53%**