# Chemical Reactions

## What is a Chemical Reaction?

A chemical reaction is when some kind of combination of chemicals rearrange their molecular or ionic structure. There are 5 kinds of chemical reactions Synthesis, Decomposition, Single Replacement, Double Replacement, and Combustion.

Between Lead (II) Nitrate and Sodium Bromide the chemical reaction is double replacement. The balanced equation looks like this Pb(NO3)2 (aq) + 2 NaBr (aq) = 2 NaNO3 (aq) + PbBr2 (s).

## Molar Mass

The set up for these numbers will be the same as the formula seen above.

269.20 g + 205.39 g = 169.94 g + 266.23 g

## Mole to Mole conversion

Moles tell how much of a certain element or compound is present. To convert moles to moles you need the moles of the fist element then how many moles of each are in the reaction and then you need to set up the formula as seen in the picture below. the moles in the reaction are found with the coefficients. I was given 11.30 moles for Pb(NO3)2 and 12.1 moles for PbBr2 and calculated the moles of the other compounds.

## Mass to Mass

Mass is the amount of matter is present and has to do with the amount of weight of the moles in grams. To convert from one element's mass to another's you first need to convert to the moles of the first element using the molar mass of the compound then the moles of the second the the moles to mass.

For this equation I was given a mass of 11.25 grams for Pb(NO3) and had to find how much mass of NaNO3 was present

## Limiting and Excess Reactants

A limiting reactant is the certain reactant that would only allow for a certain amount of the final solutions. The excess reactant is any left over of a certain initial element or compound.

For example in my reaction if my limiting reactant was PbNO3 and it had 5.25g that would give a yield of 2.59 g of PbBr2. My excess reactant would therefore be NaBr let's say there's 12.18 g of it. that would allow for the creation of 7.8g of PbBr, but since the Lead (II) Nitrate limited the reaction there is let over Sodium Bromide.

## Theoretical and Percentage Yield

The theoretical yield of this reaction would be 2.59g of Lead (II) Bromide because with the LR and ER above that is the amount we can make. Using the TY we can calculate how much percentage is actually there.

When I was given the variable of 3.62g of actual yield I divided by the theoretical yield to get a PY of 139.7%. This is technically impossible. You can't have more solution than you start with, but it is because the reaction was rushed and there wasn't enough time for all the impurities to come out, or the reaction isn't yet dry.

## The Reaction

The reaction between Lead (II) Nitrate and Sodium Bromide, two aqueous compounds, results in the creation of an aqueous compound and a solid precipitate. The compounds reacting results in a white solid powdery looking substance.