An aqueous solution containing 5.97 g of lead(II) nitrate is added to an aqueous

ID: 1057026 • Letter: A

Question

An aqueous solution containing 5.97 g of lead(II) nitrate is added to an aqueous solution containing 6.23 g of potassium chloride.

1. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.

2. What is the limiting reactant?

3. The percent yield for the reaction is 82.1%, how many grams of precipitate were recovered?

4. The percent yield for the reaction is 82.1%, how many grams of precipitate were recovered?

5. How many grams of the excess reactant remain?

Balanced equation:
Pb(NO3)2(aq) + 2 KCl(aq) = PbCl2(s) + 2 KNO3(aq)

5.97 g of lead(II) nitrate = 5.97 / 331.209 = 0.01802 Moles

6.23 g of potassium chloride = 6.23 /74.55 = 0.0835 Moles

Limiting reagent is Lead Nitrate

0.01802 Moles of lead nitrate will need 0.03604 Moles of KCl to produce 0.01802 Moles of precipitate

0.01802 moles of PbCl2 = 0.01802 x 278.106 = 5.0128 gm

Theoretical yield - 5.018 gm

Hence weight of the precipitate for 82.1 % yield = 82.1 x 5.018 / 100 = 4.12 gm

4.12 gm of preciptate will bve obtained for 82.1 % yield

Excess reagent = 0.0835 - 0.03604 = 0.04746 x 74.55 = 3.54 gm of KCl will be remaining

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