200cm3 of each of 0.1m of solutions of lead (ii) trioxonitrate (v) and hydrochloric acid were mixed. Assuming that lead (ii) chloride is completely insoluble, calculate the mass of lead (ii) chloride that will be precipitated.
Pb(NO3)2 + 2HCl ==> PbCl2 + 2HNO3
Assuming you meant 0.1M and not 0.1m,
millimols Pb(NO3)2 = mL x M = ?
millimols HCl = mL x M = ?
Determine the limiting reagent and you should see that to be HCl so that means you heave 10 millimols Pb(NO3)2 excess. Convert to mols, then grams PbCl2 = mols x molar mass PbCl2 = ?
Ther is an excess of Pb^2+ and that is a common ion. You can see how that affects the solubility of PbCl2 but I doubt you need to go through that calculations. If you want to go through it, it is done this way.
............PbCl2 ==> Pb^2+ + 2Cl^-
I...........solid......0.......0
C...........solid......x.......2x
E...........solid......x.......2x
Ksp = (Pb^2+)(Cl^-)^2
Look up Ksp for PbCl2.
(Pb^2+) = 10 millimols/400 mL = ?. Put that in for Pb^2+ in the Ksp expression above, solve for Cl^-, take half of that to give solubility of PbCl2, convert to grams PbCl2, subtract from the solid you calculated above. As I stated above, I doubt this will be significant. The difference is the amount in grams of PbCl2 that is produced.
To calculate the mass of lead (II) chloride precipitated, we need to determine the limiting reactant first. The balanced equation for the reaction between lead (II) trioxonitrate (V) and hydrochloric acid is:
Pb(NO3)2(aq) + 2HCl(aq) -> PbCl2(s) + 2HNO3(aq)
From the equation, we can see that the mole ratio of Pb(NO3)2 to PbCl2 is 1:1.
First, let's calculate the number of moles of each reactant using the given volume and concentration:
Volume of solution = 200 cm3 = 0.2 L
Number of moles of Pb(NO3)2 = concentration × volume
= 0.1 mol/L × 0.2 L
= 0.02 moles
Number of moles of HCl = concentration × volume
= 0.1 mol/L × 0.2 L
= 0.02 moles
Now, we need to compare the moles of each reactant to determine the limiting reactant. Since the stoichiometric ratio is 1:1, the reactant with fewer moles is the limiting reactant.
In this case, both reactants have the same number of moles (0.02 moles). Therefore, they are present in equimolar amounts.
Using stoichiometry, 1 mole of Pb(NO3)2 will yield 1 mole of PbCl2. Therefore, the mass of PbCl2 precipitated will be equal to the molar mass of PbCl2 times the number of moles of PbCl2.
The molar mass of PbCl2 = 207.2 g/mol (atomic mass of Pb = 207.2 g/mol)
Mass of PbCl2 = molar mass × number of moles
= 207.2 g/mol × 0.02 moles
= 4.144 g
Therefore, approximately 4.144 grams of lead (II) chloride will be precipitated.
To calculate the mass of lead (II) chloride that will be precipitated, we need to determine the limiting reactant in the reaction between lead (II) trioxonitrate (V) and hydrochloric acid. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be formed.
Let's start by writing the balanced equation for the reaction:
Pb(NO3)2 + 2HCl -> PbCl2 + 2HNO3
From the equation, we can see that one mole of lead (II) trioxonitrate (V) (Pb(NO3)2) reacts with two moles of hydrochloric acid (HCl) to produce one mole of lead (II) chloride (PbCl2) and two moles of nitric acid (HNO3).
First, let's calculate the number of moles of Pb(NO3)2 and HCl in the given volume of 0.1 M solutions:
Number of moles of Pb(NO3)2 = concentration (mol/L) x volume (L)
= 0.1 mol/L x 0.2 L
= 0.02 moles
Number of moles of HCl = concentration (mol/L) x volume (L)
= 0.1 mol/L x 0.2 L
= 0.02 moles
Next, we need to determine the stoichiometric ratio between Pb(NO3)2 and PbCl2. From the balanced equation, we can see that 1 mole of Pb(NO3)2 produces 1 mole of PbCl2.
Since the stoichiometric ratio is 1:1, the limiting reactant will be the one that has the smallest number of moles. In this case, both Pb(NO3)2 and HCl have the same number of moles (0.02 moles).
Therefore, the number of moles of PbCl2 formed will also be 0.02 moles.
Finally, we can calculate the mass of lead (II) chloride (PbCl2) using its molar mass:
Mass = number of moles x molar mass
= 0.02 moles x molar mass of PbCl2
The molar mass of PbCl2 is approximately 278.1 g/mol.
Mass = 0.02 moles x 278.1 g/mol
= 5.562 g
Therefore, approximately 5.562 grams of lead (II) chloride will be precipitated.