Question

An aqueous solution containing 9.49 g of lead(II) nitrate is added to an aqueous solution containing 5.30 g of potassium chloride.
The percent yield for the reaction is 82.6% . How many grams of precipitate is recovered? precipitate recovered:___g
How many grams of the excess reactant remain? excess reactant remaining:___g

Answer 1

The balanced reaction between lead (II ) nitrate and potassium chloride can be written as

Pb(NO3)2(aq) + 2K Cl(aq) + PbCl2(g) + 2K NO3(aq)

Note that PbCl2 in insoluble in water and will be precipitated out.

Amount of Pb(NO3)2 taken = 9.49 g

Molar mass of Pb(NO3)2 = 331.2 g/mol

Hence, number of moles of Pb(NO3)2 taken is

mass molar mass 9.49 g 331.2 g/mol i = 0.0286 mol

Mass of KCl taken = 5.30 g

Molar mass of KCl = 74.55 g/mol

Hence, number of moles of KCl taken is

5.30 g mass molar mass 74.55 g/mol i = 0.0711 mol

Note that, 1 mole of Pb(NO3)2 reacts completely with 2 moles of KCl.

Hence, the number of moles of KCl that will react with the available 0.0286 mol of Pb(NO3)2 is

2 mol KCI x 0.0286 mol Pb(NO3)2 -0.0573 mol KCI 1 mol Pb(NO3)2

Note that we have 0.0711 mol of KCl available. Hence, after completion of reaction, some of the excess KCl will be left unreacted.

Note that 1 mol of lead nitrate results in 1 mol of PbCl2 when the yield is 100% as predicted by the balanced equation.

But it is given that the yield is 82.6 %.

Hence, the number of moles of PbCl2 formed from reaction of 0.0286 mol of Pb(NO3)2 is

82.6 100 x 0.286 mol 0.0237mol

Molar mass of the precipitate PbCl2 = 278.1 g/mol

Hence, mass of precipitate formed is

0.0237mol x 278.1– 76.58 g

Hence, precipitate recovered = 6.58 g

Moles of excess KCl remaining after reaction = starting moles of KCl - moles reacted = 0.0711 mol - 0.0573 mol = 0.0138 mol

Molar mass of KCl = 74.55 g/mol

Hence, mass of KCl remaining is

0.0138 mol x 74.559, 1.03 g

Hence, excess reactant remaining = 1.03 g.