An aqueous solution containing

Question

An aqueous solution containing 7.40 g of lead(II) nitrate is added to an aqueous solution containing 5.94 g of potassium chloride.

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

balanced chemical equation:

The percent yield for the reaction is 82.4%. How many grams of the precipitate are formed?

precipitate formed:

Taking into account the percent yield, how many grams of the excess reactant (the reactant that is not limiting) remain?

excess reactant remaining:

Answer 1

Pb(NO3)₂(aq) + 2KCl(aq) ==> PbCl₂(s) + 2KNO₃(aq) ... balanced equation

The approach here is to first calculate the theoretical yield. To do this, we need to determine which reactant is limiting. After finding the theoretical yield, we will use the % yield (82.4%) to find the actual yield in grams. Finally, we will use the above information to find grams of excess reactant remaining.

Finding limiting reactant (divide moles of each reactant by corresponding coefficient in balanced equation):

Pb(NO₃)₂: 7.40 g Pb(NO₃)₂ x 1 mol / 331 g = 0.02236 mols (÷1 -> 0.02236)

KCl: 5.94 g KCl x 1 mol / 74.6 g = 0.07962 mols (÷2 -> 0.0398)

Since 002236 is less than 0.0398, Pb(NO₃)₂ is LIMITING

Theoretical yield of precipitate (PbCl₂):

0.02236 mols Pb(NO₃)₂ x 1 mol PbCl₂ / mol Pb(NO₃)₂ x 278 g /mol = 6.216 g PbCl₂ theoretical yield

Actual yield = 82.4% x 6.216 g = 5.122 g PbCl₂ (grams of precipitate)

Excess reactant (KCl) remaining:

5.122 g PbCl2 x 1 mol PbCl2 / 278 g = 0.0184 mols PbCl2 formed

0.0184 mols PbCl2 x 2 mol KCl / 1 mol PbCl2 = 0.0368 mols KCl used in the reaction

moles KCl originally present = 0.07962 moles KCl

moles KCl remaining = 0.07962 mols - 0.0368 mols = 0.0428 mols KCl remaining

grams KCl remaining = 0.0428 mols x 74.6 g /mol = 3.19 g KCl remaining