How many grams of PbBr2 will precipitate when excess FeBr2 solution is added to 47.0 mL of 0.563 M Pb(NO3)2 solution?
Pb(NO3)2(aq) + FeBr2(aq) PbBr2(s) + Fe(NO3)2(aq)
Hello kammy,
The first thing that you know about the chemical equation provided is that you know the reactants, the products, and their states (whether a substance is a solid, liquid, gas, or dissolved in an aqueous solution). You also know the molarity (M) and volume (mL) of lead (II) nitrate.
First determine the amount of substance (moles, or mol) of lead (II) nitrate from the molarity and volume of the same. For that you need to convert the unit of volume (in this case milliliters, mL) to liters, L. 47 mL is the same as 0.047 L.
Afterwards, set up an algebraic equation like the following;
Molarity = number of moles divided by the volume
0.563 M = (x mol)/(0.047 L)
Remember that molarity represented by M can be represented also by the unit mol/L.
0.563 mol/L * 0.047 L = x mol
The unit L cancels.
0.026461 = x mol
Now that you have the moles of lead (II) nitrate on one side, balance the provided chemical equation on another side. Lucky for us, the equation is already balanced. Thus, the stoichiometric coefficients for all reactants and products are 1.
Using the coefficients to represent how many moles of the reactants/products we have, we can set up the following algebraic equation;
Number of moles of lead (II) nitrate * ratio of moles of lead (II) bromide to moles of lead (II) nitrate = moles of lead (II) bromide
0.026461 mol Pb(NO3)2 * (1 mol PbBr2)/(1 mol Pb(NO3)2) = 0.026461 mol PbBr2
Now we know how many moles of PbBr2 we have: the same amount as the moles of Pb(NO3)2. This is because the coefficients of both these substances in the chemical equation are 1.
Now, we simply convert the moles of PbBr2 to grams using the periodic table (I'm gonna leave that for you to do).
I sincerely hope this helps even though I am replying after 7 years.
Hey DoctorFred,
Just wanted to say that wow ur explanation was perfect, i don't know how to thnk u. I hope I ace the midterm exam with your help.
- Kiddo
To find the number of grams of PbBr2 that will precipitate, we need to determine the limiting reactant between Pb(NO3)2 and FeBr2.
First, we need to calculate the number of moles of Pb(NO3)2 present in the solution:
Molarity of Pb(NO3)2 = 0.563 M
Volume of Pb(NO3)2 solution = 47.0 mL = 0.0470 L
Number of moles of Pb(NO3)2 = Molarity x Volume
= 0.563 M x 0.0470 L
≈ 0.0264 moles
According to the balanced chemical equation, the stoichiometric ratio between Pb(NO3)2 and PbBr2 is 1:1. This means that for every 1 mole of Pb(NO3)2, we will get 1 mole of PbBr2.
Now, let's calculate the number of moles of PbBr2:
Number of moles of PbBr2 = 0.0264 moles
Since the stoichiometric ratio between FeBr2 and PbBr2 is 1:1, we can conclude that the same number of moles of FeBr2 is required to react completely with Pb(NO3)2 to form PbBr2.
Therefore, the number of moles of PbBr2 that will precipitate is also 0.0264 moles.
To find the mass of PbBr2, we need to use its molar mass. The molar mass of PbBr2 is the sum of the atomic masses of lead (Pb) and bromine (Br), multiplied by 2 (since there are 2 bromine atoms in PbBr2):
Molar mass of PbBr2 = (atomic mass of Pb + 2 x atomic mass of Br)
= (207.2 g/mol + 2 x 79.9 g/mol)
= 207.2 g/mol + 159.8 g/mol
= 367.0 g/mol
Finally, we can calculate the mass of PbBr2:
Mass of PbBr2 = Number of moles x Molar mass
≈ 0.0264 moles x 367.0 g/mol
≈ 9.67 grams
Therefore, approximately 9.67 grams of PbBr2 will precipitate when excess FeBr2 solution is added to 47.0 mL of 0.563 M Pb(NO3)2 solution.
Here are the steps.
1. Write and balanced the equation.
2. Convert what you have to mols. mols = M x L = ?
3. Using the coefficients in the balanced equation, convert mols of what you have to mols of what you want.
4. Now convert mols of the product to grams. g = mols x molar mass.