Question

1)Write the complete equation (balanced with states) for the reaction between potassium chloride and lead (II) nitrate?

2) Which precipitate forms in the above mentioned reaction?

3) You mix 16.00 mL of 0.778 M lead (II) nitrate with 18.00 mL of 0.813 M potassium chloride, and collect and dry the precipitate. Calculate the following:

a) Moles of lead (II) nitrate used:

b)Moles of potassium chloride used:

c) Moles of precipitate expected from lead (II) nitrate:

d) Moles of precipitate expected from potassium chloride:

e) Compare parts c and d to indicate the limiting reactant.

f) What mass of precipitate do you expect to obtain from this reaction?

g) If the actual yield of the precipitate you recover is 0.916g, what is the percent yield of the precipitate?

show work please

Answer 1

1) The reaction between potassium chloride and lead(II)nitrate is,

Pb(NO₃)₂(aq) + 2KCl(aq) PbCl₂(s) + 2KNO₃(aq)

2) Lead(II)chloride is the precipitate forms in the above-mentioned reaction,

3) Given,

[Pb(NO₃)₂] = 0.778 M

Volume of Pb(NO₃)₂ = 16.00 mL x ( 1L /1000 mL) = 0.016 L

[KCl] = 0.813 M

Volume of KCl solution = 18.00 mL x ( 1 L /1000 mL) = 0.018 L

a) Moles of lead(II) nitrate used = 0.778 M x 0.016 L = 0.0124 mol of Pb(NO₃)₂

b) Moles of potassium chloride used = 0.813 M x 0.018 L = 0.0146 mol of KCl

c) Moles of precipitate expected from lead(II)nitrate

Now, using the moles of lead(II)nitrate and the mole ratio from the balanced chemical reaction, calculating the number of moles of precipitate formed,

= 0.01245 mol of Pb(NO₃)₂ x ( 1 mol PbCl₂ / 1 mol of Pb(NO₃)₂)

= 0.0124 mol of PbCl₂ precipitate expected from lead(II) nitrate

d) Moles of precipitate expected from potassium chloride

Now, using the moles of potassium chloride and the mole ratio from the balanced chemical reaction, calculating the number of moles of precipitate formed,

= 0.01463 mol of KCl x ( 1 mol PbCl₂ / 2 mol of KCl)

= 0.00732 mol of PbCl₂ precipitate expected from potassium chloride

e) From part c and part d, potassium chloride is the limiting reactant, since the precipitate expected from KCl is less in quantity than the precipitate expected from Pb(NO₃)₂

f) Converting the number of moles of precipitate to grams formed from the limiting reactant,

= 0.007317 mol of PbCl₂ x (278.1 g / 1 mol)

= 2.03 g of PbCl₂ precipitate expect to obtain.

g) Given, the actual yield of PbCl₂ precipitate = 0.916 g

We know, the formula for percent yield,

Percent yield = [Actual yield / Expected yield] x 100

Percent yield = [0.916 g / 2.03 g] x 100

Percent yield = 45.0 %