# Lead II Nitrate & Sodium Bromide

## Balancing your equation, IUPAC, and Molar Mass

This is called a double reaction. To balance your equation you first need to write it out.

Pb(NO₃)₂ + NaBr ---> Na(NO₃) + PbBr₂

To balance you need to make sure the reactants and products have the same amount of elements on each side.

Pb(NO₃)₂ + 2NaBr ---> 2Na(NO₃) + PbBr₂

Na(NO₃) needed a 2 coefficient because Pb(NO₃)₂ has 6 (NO₃). NaBr needed a 2 because we added a 2 on Na(NO₃) on the product side so we need it to be equal on the reactant side.

IPUAC name:To write the IUPAC name you basically write the reaction into words.

Ex: Lead ll Nitrate and Sodium Bromide react with Sodium Nitrate and Lead ll Bromide

Molar Mass: To find molar mass you add the masses of each element (including charges) together.

Ex: The molar mas of NaBr is 102.894. I found this by adding 22.990 (the mass of Na) and 79.904 (the mass of Br.)

MM: Pb(NO₃)₂ + 2NaBr ---> 2Na(NO₃) + PbBr₂

331.208 102.894 84.994 367.008

## Mole to Mole and Mass to Mass conversions

Mole to Mole: You start with mole A (given from the question) then multiply this by the coefficient from mole B (the desired compound.) Finally, divide the answer by the coefficient of mole A.

Ex: 12.14 x 1 (Mole of PbBr₂) / 1 (Mole of Pb(NO₃)₂ = 12.14 Mole PbBr₂

Mass to Mass: You start with Mass A (given from the question) and multiply this by 1 (Mole a) then multiply this by mole B's coefficient and then times this by the molar mass of mole B. Then divide the answer by the molar mass of mole A and the coefficient of mole A and then by 1 (Mole b).

Ex: 12.1 (Pb(NO₃)₂) x 1 x 2 (mole NaBr₂) x 102.894 g / 331.208 (G Pb(NO₃)₂) / 1 (Mole Pb(NO₃)₂) / 1 (mole NaNO₃) = 7.5 g NaBr₂

Mole-to-mole and Mass-to-mass Conversions

## Limiting and Excess Reactants

To find the limiting and excess reactants you need to do two equations. You will first start with the first compound on the reactant side. (Ex: PbNO₃)₂). Work the problem out like a mass to mass conversion. Mole B will be the first product compound (ex: Na(NO₃)). When you are finished with the first equation then start on the second one. On the second one Mass A will be the second compound on the reactant side (Ex: NaBr) and Mole B will be the same from the first question (Na(NO₃)). Once you find both answers, the smallest one is the limiting reactant. The larger answer is the excess.

Ex:

First equation answer = 6.31 g of Na(NO₃)

Second equation answer = 10 g of Na(NO₃)

12.3g Pb(NO₃) = Limiting reactant

12.3g NaBr = Excess reactant

## Theoretical and Percent Yeild

Theoretical Yield: Take the limiting reactant and use its answer. Basically do a limiting and excess problem and take the smallest answer and there it is.

Ex: 6.31g Na(NO₃) would be the Theoretical yield in the limiting and excess problem above.

Percent Yield: Take the actual yield and divide it by the theoretical yield.

Ex: 4.92 (g of Na(NO₃)) / 6.31 (g of Na(NO₃)) = 77.97%

## Real world application of Stoichiomerty

Rocket Fuel!
Stoichiometry - Rocket Fuel