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
The balanced reaction between lead (II ) nitrate and potassium chloride can be written as
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 of KCl taken = 5.30 g
Molar mass of KCl = 74.55 g/mol
Hence, number of moles of KCl taken is
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
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
Molar mass of the precipitate PbCl2 = 278.1 g/mol
Hence, mass of precipitate formed is
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
Hence, excess reactant remaining = 1.03 g.
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