Question

An aqueous solution containing 6.16 g of lead(II) nitrate is added to an aqueous solution containing 6.62 g of potassium chloride.
what is the balanced chemical equation for this reaction?
what is the limiting reactant?The percent yield for the reaction is 79.7% . How many grams of precipitate is recovered?
How many grams of the excess reactant remain?

Answer 1

Solution.

The balanced chemical equation for this reaction:

$Pb(NO3)2 + 2KCl = PbCl2 + 2KNO3$

Now we find the limiting reagent through the amount of substance:

$n(Pb(NO3)2) = \frac{m(Pb(NO3)2)}{M(Pb(NO3)2)}$

$n(KCl) = \frac{m(KCl)}{M(KCl)}$

n(Pb(NO3)2) = 0.02 mole

n(KCl) = 0.09 mole

So, Pb(NO3)2 is limiting reagent.

Now we should find m(PbCl2). Since n(Pb(NO3)2) is equal to n(PbCl2), so we find m(PbCl2).

$m(PbCl2) = n(PbCl2) \times M(PbCl2)$

m(PbCl2, theor.) = 5.56 g

The yield of the reaction product is the ratio of the mass of the substance obtained in the experiment (practical) to the mass of the precipitate obtained in theory (theoretical).

$\eta = \frac{m(PbCl2, pract.)}{m(PbCl2, theor.)} \times 100 \%$

To find the mass of sediment in the experiment, we express the practical mass from the formula.

$m(PbCl2, pract.) = \frac{\eta \times m(PbCl2, theor.)}{100 \%}$

m(PbCl2, pract.) = 4.43 g

Answer:

The balanced chemical equation for this reaction:

$Pb(NO3)2 + 2KCl = PbCl2 + 2KNO3$

Pb(NO3)2 is limiting reagent.

m(PbCl2, pract.) = 4.43 g