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 0.0350 M Fe(II)?

Answer 1

The balanced reaction is as follows:

4Fe(OH)⁺(aq) + 4OH^-(aq) + O₂(g) + 2H₂O(l) 4Fe(OH)₃(s)

Concentration of Fe(OH)⁺ solution = 0.0350 M

Volume of Fe(OH)⁺ solution = 50.0 mLx( 1 L/1000 mL) = 0.05 L

Determine the number of moles of Fe(OH)⁺ in the aqueous solution as follows:

The formula to determine molarity is as follows:

Molarity = Number of moles / L of solution

Rearrange the formula for number of moles as follows:

Number of moles = Molarity x L of solution

Number of moles = 0.0350 M x 0.05 L

Mol Fe(OH)⁺ = 0.00175 mol Fe(OH)⁺

Use the moles of Fe(OH)⁺ and the balanced chemical reaction to determine the number of moles of oxygen consumed as follows:

Mol O₂ = 0.00175 mol Fe(OH)⁺ x (1 mol O₂ / 4 mol Fe(OH)⁺)

Mol O₂ = 0.000438 mol O₂

Convert moles of oxygen to grams as follows:

= 0.000438 mol O₂ x ( 32 g O₂ / 1 mol O₂)

=0.014 g O₂