Question

please help with the excess reactant remaining (at bottom of second picture)

Attempt An aqueous solution containing 7.42 g of lead(II) nitrate is added to an aqueous solution containing 6.44 g of potassium chloride. Enter the balanced chemical equation for this reaction. Be sure to include all physical states. balanced chemical equation: Pb(NO,), (aq) +2KCI(aq) PbCl,(s) + 2KNO, (aq) What is the limiting reactant? lead(II) nitrate potassium chloride The percent yield for the reaction is 89.3 %. How many grams of precipitate is recovered? precipitate recovered: 5.56 How many grams of the excess reactant remain?

Answer 1

1)

Balanced chemical equation is:

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

2)

Molar mass of Pb(NO3)2,

MM = 1*MM(Pb) + 2*MM(N) + 6*MM(O)

= 1*207.2 + 2*14.01 + 6*16.0

= 331.22 g/mol

mass(Pb(NO3)2)= 7.42 g

use:

number of mol of Pb(NO3)2,

n = mass of Pb(NO3)2/molar mass of Pb(NO3)2

=(7.42 g)/(3.312*10^2 g/mol)

= 2.24*10^-2 mol

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

mass(KCl)= 6.44 g

use:

number of mol of KCl,

n = mass of KCl/molar mass of KCl

=(6.44 g)/(74.55 g/mol)

= 8.638*10^-2 mol

1 mol of Pb(NO3)2 reacts with 2 mol of KCl

for 2.24*10^-2 mol of Pb(NO3)2, 4.48*10^-2 mol of KCl is required

But we have 8.638*10^-2 mol of KCl

so, Pb(NO3)2 is limiting reagent

Answer: lead(II) nitrate

3)

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

According to balanced equation

mol of PbCl2 formed = (1/1)* moles of Pb(NO3)2

= (1/1)*2.24*10^-2

= 2.24*10^-2 mol

use:

mass of PbCl2 = number of mol * molar mass

= 2.24*10^-2*2.781*10^2

= 6.23 g

% yield = actual mass*100/theoretical mass

89.3= actual mass*100/6.23

actual mass=5.563 g

Answer: 5.56 g

4)

According to balanced equation

mol of KCl reacted = (2/1)* moles of Pb(NO3)2

= (2/1)*2.24*10^-2

= 4.48*10^-2 mol

mol of KCl remaining = mol initially present - mol reacted

mol of KCl remaining = 8.638*10^-2 - 4.48*10^-2

mol of KCl remaining = 4.158*10^-2 mol

Molar mass of KCl,

MM = 1*MM(K) + 1*MM(Cl)

= 1*39.1 + 1*35.45

= 74.55 g/mol

use:

mass of KCl,

m = number of mol * molar mass

= 4.158*10^-2 mol * 74.55 g/mol

= 3.10 g

Answer: 3.10 g