Question

39

What is the theoretical yield (in g of precipitate) when 15.8 mL of a 0.572 M solution of iron(III) chloride is combined with 16.8 mL of a 0.687 M solution of lead(II) nitrate?

Answer 1

Given the volume of Iron(III) chloride, FeCl3 = 15.8 mL = (15.8mL)x(1L / 1000 mL) = 0.0158 L

Concentration of FeCl3 = 0.572 M

Hence moles of FeCl3 = MxV = 0.572M x 0.0158L = 0.00904 mol

Given the volume of lead(II) nitrate, Pb(NO3)2 = 16.8 mL = (16.8mL)x(1L / 1000 mL) = 0.0168 L

Concentration of Pb(NO3)2 = 0.687 M

Hence moles of Pb(NO3)2 = MxV = 0.687M x 0.0168L = 0.0115 mol

When iron(III) chloride reacts with lead(II) nitrate, the following reaction occurs and PbCl2 precipitates.

2FeCl3 + 3Pb(NO3)2 ------> 2Fe³⁺(aq) + 6NO₃^-(aq) + 3PbCl2 (s -ppt)

2 mol 3 mol 2 mol 6 mol 3 mol

Stoichoimetric ratio of FeCl3 to Pb(NO3)2 = 2/3 = 0.667

Given molar ratio of FeCl3 to Pb(NO3)2 = 0.00904 / 0.0115 = 0.786, which is more than the stoichiometric ratio of 0.667. Hence FeCl3 is present in excess and  Pb(NO3)2 is the limiting reagent and the later will be exhausted first and will determine the amount of the products formed.

3 mol of Pb(NO3)2 forms the precipitation of PbCl2 = 3 mol PbCl2

0.0115 mol of Pb(NO3)2 that will form the precipitation of PbCl2

= [3 mol PbCl2 / 3 mol Pb(NO₃)₂]x [0.0115 mol Pb(NO₃)₂]

= (3/3) x 0.0115 mol PbCl2 = 0.0115 mol PbCl2

= [0.0115 mol PbCl2]x[278 g PbCl2/ 1 mol PbCl2] = 3.20 g (answer)