What mass of lead (II)chloride would precipitate if a solution containing 8,28 g of lead (II) ions(Pb2+) reacts with excess HCl?
Pb^2+ + HCl ==> PbCl2 + 2H^+
mols Pb^2+ = grams/atomic mass
Convert mols Pb^2+ to mols PbCl2 (Use the coefficients to do this)
Convert mols PbCl2 to grams. g = mols x molar mass.
To determine the mass of lead (II) chloride that would precipitate in the reaction, you need to consider the balanced chemical equation for the reaction between lead (II) ions and hydrochloric acid (HCl):
Pb2+ (aq) + 2Cl- (aq) → PbCl2 (s)
From the balanced equation, you can see that one mole of lead (II) ions reacts with two moles of chloride ions to produce one mole of lead (II) chloride.
Now, let's calculate the number of moles of lead (II) ions that are present in the solution using the given mass of lead (II) ions:
Mass of lead (II) ions (Pb2+) = 8.28 g
Molar mass of Pb = 207.2 g/mol (atomic mass of Pb = 207.2 g/mol)
Moles of Pb2+ = mass of Pb2+ / molar mass of Pb
Substituting the values:
Moles of Pb2+ = 8.28 g / 207.2 g/mol ≈ 0.04 mol
According to the balanced equation, the ratio between Pb2+ and PbCl2 is 1:1. Therefore, the moles of lead (II) chloride formed will also be 0.04 mol.
To calculate the mass of lead (II) chloride formed, we need to know its molar mass. The molar mass of lead (II) chloride (PbCl2) can be calculated as follows:
Molar mass of PbCl2 = atomic mass of Pb + 2 * atomic mass of Cl
= 207.2 g/mol + 2 * 35.45 g/mol (atomic mass of Cl)
= 207.2 g/mol + 70.90 g/mol
= 278.10 g/mol
Now, we can calculate the mass of lead (II) chloride:
Mass of PbCl2 = moles of PbCl2 * molar mass of PbCl2
= 0.04 mol * 278.10 g/mol
= 11.12 g
Therefore, approximately 11.12 grams of lead (II) chloride would precipitate if a solution containing 8.28 g of lead (II) ions reacts with excess HCl.