Question

If a 5.23g of lead chloride is added to a 51.29mL sample of a 0.305M silver nitrate solution how much silver chloride could be formed?

PbCl2(s)+2AgNO3(aq)---> 2AgCl(s)+ Pb(NO3)(aq)

Answer 1

Molar mass of PbCl2 = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

mass of PbCl2 = 5.23 g

we have below equation to be used:

number of mol of PbCl2,

n = mass of PbCl2/molar mass of PbCl2

=(5.23 g)/(278.1 g/mol)

= 1.881*10^-2 mol

volume of AgNO3, V = 51.29 mL

= 5.129*10^-2 L

we have below equation to be used:

number of mol in AgNO3,

n = Molarity * Volume

= 0.305*0.0513

= 1.564*10^-2 mol

we have the Balanced chemical equation as:

PbCl2 + 2 AgNO3 ---> 2 AgCl + Pb(NO3)2

1 mol of PbCl2 reacts with 2 mol of AgNO3

for 1.881*10^-2 mol of PbCl2, 3.761*10^-2 mol of AgNO3 is required

But we have 1.564*10^-2 mol of AgNO3

so, AgNO3 is limiting reagent

we will use AgNO3 in further calculation

Molar mass of AgCl = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

From balanced chemical reaction, we see that

when 2 mol of AgNO3 reacts, 2 mol of AgCl is formed

mol of AgCl formed = (2/2)* moles of AgNO3

= (2/2)*1.564*10^-2

= 1.564*10^-2 mol

we have below equation to be used:

mass of AgCl = number of mol * molar mass

= 1.564*10^-2*1.434*10^2

= 2.242 g

Answer: 2.24 g

Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know