Question

An industrial vat contains 650 grams of solid lead(II) chloride formed from a reaction of 870 grams of lead(II) nitrate with excess hydrochloric acid. This is the equation of the reaction: 2HCl + Pb(NO3)2 → 2HNO3 + PbCl2. What is the percent yield of lead(II) chloride? The percent yield of lead chloride is %.

Answer 1

Answer:

88.98 %.

Explanation:

• From the balanced equation:

2HCl + Pb(NO₃)₂ → 2HNO₃ + PbCl₂

• It is clear that 1.0 mole of Pb(NO₃)₂ reacts with 2.0 moles of HCl to produce 1.0 mole of PbCl₂ and 2.0 moles of HNO₃.

• The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100.

• The actual yield of lead(II) chloride (PbCl₂) = 650 g.

• Now, we need to calculate the calculated yield of lead(II) chloride (PbCl₂).

• We need to calculate the no. of moles (n) of lead(II) nitrate (Pb(NO₃)₂) (870 grams) using the relation: n = mass / molar mass.

• n of lead(II) nitrate (Pb(NO₃)₂) = mass / molar mass = (870 g) / (331.2 g/mol) = 2.63 mol.

• Since HCl is in excess, the limiting reactant is lead(II) nitrate (Pb(NO₃)₂).

Using cross multiplication:

1.0 mole of Pb(NO₃)₂ produces → 1.0 mole of PbCl₂, from the stichiometry.

∴ 2.63 mole of Pb(NO₃)₂ produces → 2.63 mole of PbCl₂.

• The mass of PbCl₂ produced (the calculated yield) = n x molar mass = (2.63 mol) (278.1 g/mol) = 730.52 g.

∴ The percent yield % of lead(II) chloride (PbCl₂) = [(actual yield) / (calculated yield)] x 100 = [(650 g) / (730.52)] x 100 = 88.98 %.

Answer 2

Answer:

89%

Explanation:

Should be calculated to two significant figures so i assume its 89%