# Stoichiometry

## The reaction We'll be using

Lead(II) Nitrate and Aluminum

The Roman numeral after lead is 2, and it is used because Lead is part of special types of metals in which it is difficult to find the subscript so the 2 is given to you as a mean to save time.

## Type of Reaction?

The reaction we have here is single replacement since we're dealing with 3 elements and when 2 of them switch, it leaves one of the elements out because it is a odd number of elements.

## Balanced Equation/Reaction

2Al+3 Pb(NO3)2>> 2Al(NO3)3+3 Pb

When balancing a reaction you have to make sure the number of atoms on the left equals the number of atoms on the right, and you do this by looking at the original equation and then adding the correct Subscripts and Coefficients in order to make it even.

## IUPAC; The names of the reactants

Lead Nitrate and Aluminum yields Aluminum Nitrate and Lead.

## Molar mass for each reactant and product

Al+Pb(NO3)2>>>Al(NO3)3+Pb

26.982+269.204=212.994+207.2

The Molar mass is the grams per mole, and it is found by looking at the periodic table and using the subscripts as a multiplier of the number on the periodic table.

Balancing Chemical Equations

## Mole to mole conversions

To start off in finding the mass you have to start with the given number in the question, which in this case, will be 12.23g of Lead Nitrate. Then we look at the subscripts and on the bottom we put 3 Moles of Lead nitrate and then 2 Moles of aluminum on top, because our goal is to find the grams of aluminum. We then multiply all of the top then divide by the bottom and you'll get 8.1533 moles of Al.

## Mass to Mass conversions

You start with the given which will be 12.1g of Lead Nitrate. Then you put the molar mass of lead nitrate on the bottom and one mole of lead nitrate on top. Next you use the Coefficients and have 3 moles of Lead Nitrate on the bottom then 2 moles of aluminum on top. Lastly put 1 mole of aluminum on the bottom, the molar mass of aluminum on top, then multiply everything on top and divide by the bottom to get an answer of .8088 Al

## Limiting and excess reactant

The limiting reagent (or limiting reactant) in a chemical reaction is the substance which is totally consumed when the chemical reaction is complete. The reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed.To find it, we follow this diagram. Try it yourself!! The given will be 12g of Lead Nitrate and the correct answers will be 12.3g of Aluminum nitrate as the excess and 12g of Lead Nitrate as the Limiting.

## Theoretical yield

The theoretical yield for a reaction is calculated based on the limiting reactant.

To find it,you use the limiting reactant and follow the diagram, then use the answer as the given for the 2nd equation. You use the same steps for the 2nd equation as you do for the 1st. After you solve the 2nd equation and get an answer, you compare both the 1st and 2nd answer, and the smaller number is your theoretical yield. The answer should be 6.3295.

STOICHIOMETRY - Limiting Reactant & Excess Reactant Stoichiometry & Moles

## Percent yield

When you divide actual yield by theoretical yield you get a decimal percentage known as the percent yield of a reaction. In This case, we would divide 6.329 by 8.53(actual yield) and then multiply by 100 for an answer of 74.09 percent.
Stoichiometry: Chemistry for Massive Creatures - Crash Course Chemistry #6