Question

You mix 24.0 mil. of 0.276 M FeCl with 43.5 ml. of 0 545 M NaOH What mass of Fe(Ol)s (in grams) will preciptate from this reaction micture? Mass Oae of the reactants (FeCl or NaOH) is present in a stoichsiometric excess What is the molar concentration of the excess atant emaining in solution afer Fe(OHl)y has beee precipitatod? Asume dhe volames are addhave Try Another Version 2hem alttempts remaining

Answer 1

The balanced chemical equation for the reaction is

FeCl₃(aq) + 3NaOH(aq) $\rightarrow$ Fe(OH)₃ (s) + 3NaCl

Concentration of FeCl₃(aq) =0.276 mol/L

Volume of FeCl₃(aq) =24mL

No. of moles of FeCl₃ in 24 mL solution = [0.276 mol/1000mL] x 24 mL (1L= 1000mL)

=0.006624 moles

Concentration of NaOH(aq) =0.545 mol/L

Volume of  NaOH (aq) =43.5 mL

No. of moles of  NaOH in 43.5 mL solution = [0.545 mol/1000mL] x 43.5 mL (1L= 1000mL)

=0.02371 moles

From the reaction stoichiometry,

1 mole of FeCl₃ requires 3 moles of NaOH

So, No. of moles of NaOH required to completely react with 0.006624 moles of FeCl₃ =3 x 0.006624 moles

=0.01987 moles

Now, available moles of NaOH (0.02371 moles) is more than required moles(0.01987 moles), therefore NaOH will not get exhausted during the reaction.

The reactant to get exhausted is FeCl_3, therefore FeCl₃ is the limiting reagent.

1 mole of FeCl₃ produces 1 mole of Fe(OH)₃ ( from the balanced equation)

Therefore,0.006624 moles of FeCl₃ will produce 0.006624 moles of Fe(OH)₃

Molar mass of Fe(OH)₃ =162.2 g/mol

Mass of 0.006624 moles of Fe(OH)₃=0.006624 moles x 162.2 g/mol

= 1.074 g

Mass of  Fe(OH)₃ precipitated =1.074 g

The reactant in stoichiometric excess is NaOH.

No. of moles of NaOH in excess= No. of moles of NaOH present - No. of moles of NaOH consumed

=0.02371 moles-0.01987 moles

=0.00384 moles

Total volume of solution=( 24+43.5)mL =67.5 mL= 0.0675 L

Concentration of excess NaOH left in solution=( No. of moles of NaOH in excess) / (Total volume of solution in L)

=(0.00384 moles) / 0.0675 L

= 0.0569 mol/L

Concentration of excess NaOH left in solution =0.0569 M