the heat of precipitation of iron (III) hydroxidde, Fe (OH) is -10kJ. When a 2mol dm^-3 sodium hydoxide, NaOH solution was mixed together with a solutionof iron

Question

the heat of precipitation of iron (III) hydroxidde, Fe (OH)₃ is -10kJ. When a 2 mol dm^-3 sodium hydoxide, NaOH solution was mixed together with a solutionof iron (III) chloride, FeCI₃, 200 J of heat was released. Calculate the volume of sodium hydroxide, NaOH solution that used.

Answer 1

Looking at the reaction taking place we have 3NaOH(aq) + FeCl₃(aq) ==> 3NaCl(aq) + Fe(OH)₃(s)

From the observation that 200 J of heat was released in the reaction, one can calculate the moles of Fe(OH)₃ produced.

200 J x 1 kJ/1000 J = 0.2 kJ heat released.

0.2 kJ x 1 mole Fe(OH)₃/10 kJ = 0.02 moles Fe(OH)₃ produced

From the balanced equation, one can now find the number of moles of NaOH needed to produce 0.02 moles Fe(OH)₃.

0.02 moles Fe(OH)₃ x 3 moles NaOH/mole Fe(OH)₃ = 0.06 moles of NaOH was used (assuming FeCl₃ is not limiting)

From moles of NaOH, one can easily calculate the volume used.

2 moles NaOH/dm³ (x dm³) = 0.06 moles

x = 0.03 dm³ = 0.03 L = 30 mls NaOH used