Fe(ii) can be precipitated from a slightly basic aqueous solution by bubbling oxygen through the solution, which converts fe(ii) to insoluble fe(iii): $$4fe(oh)+(aq)+4oh−(aq)+o2(g)+2h2o( l) 4fe(oh)3(s) how many grams of o2 are consumed to precipitate all of the iron in 50.0 ml of 0.0850 m fe(ii)?

The amount of O2 consumed is 0.034 g.

The equation of the reaction is; 4Fe(OH^+)(s) + 4OH^-(aq) + O2(g) + 2H2O(l) >  4Fe(OH)3(s)

We can obtain the amount of Fe^2+ reacted as follows;

number of moles = concentration × volume

concentration = 0.0850  M

volume = 50.0 ml  or 0.05 L

number of moles of  Fe^2+ = 0.0850  M × 0.05 L

number of moles of  Fe^2+ = 0.00425 moles

Fom the reaction equation;

4 moles of Fe^2+ reacts with 1 mole of oxygen

0.00425 moles reacts with 0.00425 moles × 1 mole/4 moles

= 0.00106 moles O2

Mass of O2 consumed = 0.00106 moles O2 × 32 g/mol = 0.034 g

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0.0340 g O2

Step 1. Write the balanced chemical equation

4Fe(OH)^(+) + 4OH^(-) + O2 + 2H2O → 4Fe(OH)3

Step 2. Calculate the moles of Fe^(2+)

Moles of Fe^(2+) = 50.0 mL Fe^(2+) × [0.0850 mmol Fe^(2+)/1 mL Fe^(2+)]

= 4.250 mmol Fe^(2+)

Step 3. Calculate the moles of O2

Moles of O2 = 4.250 mmol Fe^(2+) × [1 mmol O2/4 mmol Fe^(2+)]

= 1.062 mmol O2

Step 4. Calculate the mass of O2

Mass of O2 = 1.062 mmol O2 × (32.00 mg O2/1 mmol O2) = 34.0 mg O2

= 0.0340 g O2

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