Question

If we have 50 mL of a 1.0M sodium hydroxide solution and 50 mL of a 0.20 M iron (III) nitrate solution, what is the concentration of ions in each solution? Write the chemical, complete ionic and net ionic equations for the reaction. Chemical: Complete lonic: Net Ionic: What volume of 1.0M NaOH is required to precipitate all the Fe ions from 50. mL of a 0.20 M Fe(NO) solution? What mass of iron (II) hydroxide precipitate can be produced by reacting 50, mL of 0.2 M Fe(NOs) with 50. mL of 1.0 M NaOH? What mass of precipitate did you recover (actual yiel d)? What is your percent yield?

Answer 1

loro lo so ml of 10M NooH - 50 xl.o a 0.05 moles So, Brath & Co os moles 100 (rat] 2005 IM OH = 0.05 moles [OH - IM. 30 m Loq 0.20M Fe (N03/2 250 000 & odo moles Pezt a 6. 10 moles The 34] 0.209. No - 3x 0.10 - (0.30 mokes) [Noz] 0.50M LOOD Chemical equation & Fe (NO) last 3 Naonlay , FeCOH) (s) + 3 Nang laq) Complete sonic Petlagt 3 Notalt 3 Natalt 3 onlagt reconst 3 Natag & 3N03 Net sonier Rest +304 ) PeloH) (s) (aq) (aq)

Volume of NaOH required:

Moles of Fe³⁺ = 0.01 mol

Moles of OH- required = 3*0.01 = 0.03 mol

Moles of NaOH to be added = 0.03 mol

Volume of 1 M NaOH to be added = 0.03/1 M = 0.03 L

= 30 ml

So, limiting is Fe(NO₃)₃

Mass of Fe(OH)₃ precipitated;

1 mol of Fe³⁺ = 1 mol of Fe(OH)₃

0.010 mol of Fe³⁺ = 0.010 mol of Fe(OH)₃

mass of precipitates = 0.010 mol * 106.867 g/mol

= 1.069 g