Question

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)+2h2​o(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)?

Answer 1

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