Question

An aqueous solution containing 6.60 g of lead(II) nitrate is added to an aqueous solution containing 6.82 g of potassium chloride to generate solid lead(II) chloride and potassium nitrate, Write the balanced chemical equation for this reaction. Be sure to include all physical states. equation: What is the limiting reactant? lead(II) nitrate O potassium chloride The percent yield for the reaction is 80.3%, how many grams of precipitate were recovered? mass: How many grams of the excess reactant remain? mass:

Answer 1

Pb(NO3)2(aq) +2 KCl(aq) ----> PbCl2(s) +2 KNO3(aq)
is the balanced reaction.TO find out limiting reactant
we need # moles of each reactant present

we have mass of Pb(NO3)2 = 6.6 g
mass of KCl = 6.82 g

molar mass of Pb(NO3)2 = [1*207.2+ 2*14.01+6*16]g/mol =331.21 g/mol

Molar mass of KCl = [1*39.10 + 1*35.45]g/mol =74.55 g/mol

# moles of Pb(NO3)2 present = mass of Pb(NO3)2/molar mass=
6.6 g/331.21 g/mol = 0.0199 mol

# moles of KCl present = 6.82 g/74.55 g/mol = 0.09148 mols

mole ratio Pb(NO3)2:KCl = 1:2
hence
0.09148 mols KClneed [1/2*0.0915 mols mol ]Pb(NO3)2

= 0.04574 mols

But 0.0199 mol PB(NO3)2 is present.Which is fewer than what actually needed
for KCl
Hence
limiting reactant is Pb(NO3)2
*************************
moles of product formed depends on moles of limiting reactant
present

Moles of PbCL2 formed = moles of Pb(NO3)2 present [since mole ratio is 1:1]
= 0.0199 mols
Mass of PbCL2 formed = moles of PbCl2 *molar mass of PbCL2
= 0.0199 mols*278.11 mols/l = 5.53 g

This is the theoretical yield of product.

But percent yield = 80.3%
So actual yield = theoretical yield*% yield
= 5.53g *0.803 = 4.44 g

mass = 4.44g
***************

Since KCl is excess,
moles of KCL reacted = 0.0199 mole *2 [ since mole ratio Pb(NO3)2:KCl=1:2]
=0.0398 mols
Moles of KCl left over = 0.09148 mols-0.0398 mols = 0.05168 mols

Hence mass of KCl left over = moles of KCL left over*molar mass of KCl
= 0.05168 mols*74.55 g/mol =3.853 g
****************************
Hope it is helpful.Kindly rate :)