Lead II Nitrate & Sodium Bromide

A helpful guide to learning stoichiomitry

By: Alyssa Rice

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.


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