Question

In a laboratory experiment, 2.0 g of aqueous lead (II) nitrate was mixed with 2.0 g of aqueous sodium chloride. Upon completion of the experiment, 0.65 g of lead (II) chloride were obtained.

a) Provide the balanced equation.

b) What is the limiting reactant?

c) What is the theoretical yield?

d) What is the percent yield?

Answer 1

a)

Pb(NO3)2(aq) + 2NaCl(aq) ---- > PbCl2(s) + 2NaNO3(aq)

b)

given that

2.0 g of aqueous lead (II) nitrate

2.0 g of aqueous sodium chloride

Number of moles = amount in g / molar mass

lead (II) nitrate= 2.0 g/ 331.2 g/mol

= 0.00604 mole lead (II) nitrate

sodium chloride = 2.0 g/ 58.44 g/mol

= 0.0342 moles sodium chloride

0.00604 mole lead (II) nitrate *2 mole NaCl /1 mole lead (II) nitrate

=0.01208 mole NaCl is used

lead (II) nitrate is limiting agent which consumed completely

The limiting reagent (or limiting reactant) is a substance that limits the amount of product in the reaction. This limiting agent is completely consumed in the reaction.

c) 0.00604 mole lead (II) nitrate *1 mole lead (II) chloride /1 mole lead (II) nitrate

= 0.00604 lead (II) chloride

Amount in g = number of moles * molar mass

= 0.00604 lead (II) chloride*278.1 g/mol

= 1.68 g lead (II) chloride is theoretical yield

d)

0.65 g of lead (II) chloride

Percentage yield = observed yield / theoretical yield*100

= 0.65 g/1.68 g*100

= 38.7%