Question

An aqueous solution containing 5.26 g of lead(II) nitrate is added to an aqueous solution containing 7.33 g of potassium chloride.

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

balanced chemical equation:

What is the limiting reactant?

lead(II) nitrate

potassium chloride

The percent yield for the reaction is 84.1%84.1%. How many grams of the precipitate are formed?

precipitate formed:

How many grams of the excess reactant remain?

excess reactant remaining:

Answer 1

1. balanced chemical equation;

Pb(NO3)2(aq) + 2KCl(aq) --> PbCl2(s) + 2KCl(aq)

(331.2)     (74.55 ) ( 278.1 ) ..... these are molar masses

(5.26) (7.33) .......these are given gram masses of reactants

2. liminting agent is Pb(NO3)2

since from above balanced equation,

331.2g of lead nitrate uses =74.55g of pot. chloride

5.26g of lead nitrate will use = 74.55/331.2 *5.26 g of pot. chloride

= 1.183 g of potassium chloride

but the given mass of KCl is 7.33g which is more than the amount to be used that is 1.183g

excess reagent left = 7.33 - 1.183 = 6.147g

therefore its is excess reagent and lead nitrate will be completely consumed so it is limiting reagent.

3.to calculate mass of limiting reagent lead nitrate;

since limiting agent decides the amount of product formed,  from the above balanced equation we have

moles of lead nitrate =gram mass /molar mass = 5.26/331.2 = 0.015 moles

since from balanced equation 1 mole of lead nitrate give 1 mole of lead chloride

therefore, 0.015 moles of lead nitrate gives 0.015 moles of PbCl2

now to calculate mass of PbCl2 = moles of PbCl2 *  molar mass of PbCl2

= 0.015 * 278.1 = 4.17g

4. excess reagent alredy calculated in point 2 that is 6.147g