What mass of precipitate will form if 1.50L of concentrated Pb(ClO3)2 is mixed with 0.200L of 0.280M of NaI? Assume the reaction is complete
You must assume there is enough Pb(ClO3)2 to use all of the NaI.
Pb(ClO3)2 + 2NaI ==> PbI2 + 2NaClO3
mols NaI = M x L = ?
Convert mols NaI to mols PbI2 using the coefficients in the balanced equation.
Convert mols PbI2 to g. g = mols x molar mass.
To determine the mass of precipitate formed, we need to identify the balanced chemical equation for the reaction between Pb(ClO3)2 and NaI. From the given compounds, we can write the balanced equation as follows:
Pb(ClO3)2 + 2NaI -> PbI2 + 2NaClO3
According to the stoichiometry of this balanced equation, 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2.
Step 1: Calculate the number of moles of NaI:
Molarity of NaI = 0.280 M
Volume of NaI solution = 0.200 L
Number of moles of NaI = Molarity x Volume
= 0.280 mol/L x 0.200 L
= 0.056 mol
Step 2: Use mole ratio to determine the number of moles of PbI2:
According to the balanced equation, 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2.
Number of moles of PbI2 = 1/2 x Number of moles of NaI
= 1/2 x 0.056 mol
= 0.028 mol
Step 3: Calculate the mass of PbI2:
The molar mass of PbI2 is the sum of the atomic masses of Pb (207.2 g/mol) and I (126.9 g/mol) multiplied by 2 (since there are two I atoms):
Molar mass of PbI2 = (207.2 g/mol + 126.9 g/mol) x 2
= 334.1 g/mol
Mass of PbI2 = Number of moles x Molar mass
= 0.028 mol x 334.1 g/mol
= 9.32 g
Therefore, the mass of precipitate (PbI2) that will form is 9.32 grams.
To calculate the mass of precipitate formed, we need to determine the limiting reagent in the reaction. The limiting reagent is the reactant that is completely consumed, limiting the amount of product formed.
Let's start by writing the balanced chemical equation for the reaction between lead(II) perchlorate (Pb(ClO3)2) and sodium iodide (NaI):
Pb(ClO3)2 + 2NaI -> PbI2 + 2NaClO3
From the equation, we can see that 1 mole of Pb(ClO3)2 reacts with 2 moles of NaI to produce 1 mole of PbI2.
First, let's calculate the number of moles of Pb(ClO3)2:
The volume of Pb(ClO3)2 is given as 1.50L, but we need to convert it to moles using the molarity (concentration). The molarity (M) is given by moles of solute divided by liters of solution.
Moles of Pb(ClO3)2 = volume (in liters) x concentration (in mol/L)
Moles of Pb(ClO3)2 = 1.50L x (concentration of Pb(ClO3)2)
Next, let's calculate the number of moles of NaI:
The volume of NaI is given as 0.200L, and the concentration (molarity) is given as 0.280M.
Moles of NaI = volume (in liters) x concentration (in mol/L)
Moles of NaI = 0.200L x 0.280M
Now, we need to determine the limiting reagent. The moles of Pb(ClO3)2 and NaI are in a 1:2 ratio according to the balanced equation. Therefore, we need to compare the number of moles of Pb(ClO3)2 with half the number of moles of NaI:
Moles of Pb(ClO3)2 / Moles of NaI = (moles of Pb(ClO3)2) / (0.5 * moles of NaI)
If the ratio is less than 1, Pb(ClO3)2 is limiting. If the ratio is greater than 1, NaI is limiting.
Once we determine the limiting reagent, we can calculate the moles of PbI2 using the balanced equation. Finally, we can convert the moles of PbI2 to mass using the molar mass of PbI2.
I hope this explanation helps you understand the process of finding the limiting reagent and calculating the mass of precipitate formed in a chemical reaction.