Stoichiometry
0.083grams of lead nitrate, Pb(NO3)2, in solutions mixed with 0.300g of sodium iodide in solution. This results in the formation of a yellow precipitate of lead iodide. Calculate the weight of the precipitate.
Pb(NO3)2 + 2NaI ==> PbI2 + 2NaNO3
Convert 0.083 g Pb(NO3)2 to moles. #moles = grams/molar mass
Convert moles Pb(NO3)2 to moles PbI2 using the coefficients in the balanced equation which I wrote.
Convert moles PbI2 to grams. grams = moles x molar mass.
the 0.300g is not needed?
Yes, it IS needed but I overlooked it. However, there are 0.002 moles of NaI which is more than enough to react with all of the Pb(NO3)2 (0.00025 moles); therefore, the Pb(NO3)2 is the limiting reagent and the problem is worked exactly as I showed it. That is, the PbI2 formed will be the result of the 0.00025 moles (0.083 g) Pb(NO3)2.
To calculate the weight of the precipitate formed in the reaction between lead nitrate (Pb(NO3)2) and sodium iodide (NaI), we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, let's find the number of moles for each compound:
1. Lead Nitrate (Pb(NO3)2):
- Mass: 0.083 grams
- Molar mass: 1 * (207.2 g/mol) + 2 * (14.01 g/mol) + 6 * (16.00 g/mol) = 331.2 g/mol
- Moles = Mass / Molar mass
Moles of Pb(NO3)2 = 0.083 g / 331.2 g/mol
2. Sodium Iodide (NaI):
- Mass: 0.300 grams
- Molar mass: 22.99 g/mol + 1.01 g/mol = 24.00 g/mol
- Moles = Mass / Molar mass
Moles of NaI = 0.300 g / 24.00 g/mol
Now, let's find the stoichiometric ratio between the two compounds. From the balanced chemical equation, we can see that one mole of Pb(NO3)2 reacts with two moles of NaI to form one mole of lead iodide (PbI2).
Using the mole ratio, we can determine the theoretical amount of PbI2 that can be formed if all the reactants are used up. We will use the limiting reactant for this calculation.
1. Convert moles of Pb(NO3)2 to moles of PbI2:
Moles of PbI2 = Moles of Pb(NO3)2 * (1 mole of PbI2 / 1 mole of Pb(NO3)2)
2. Determine the mass of PbI2 formed:
Mass of PbI2 = Moles of PbI2 * Molar mass of PbI2
Finally, we can calculate the weight of the precipitate (PbI2). Let's put all the values together:
Moles of Pb(NO3)2 = 0.083 g / 331.2 g/mol
Moles of NaI = 0.300 g / 24.00 g/mol
Moles of PbI2 = Moles of Pb(NO3)2 * (1 mole of PbI2 / 1 mole of Pb(NO3)2)
Mass of PbI2 = Moles of PbI2 * Molar mass of PbI2
Using these calculations, you can find the weight of the precipitate formed.