Question

How many grams of O2 are required to precipitate all the iron in 75 mL of 0.090 M Fe2+?

Answer 1

mass O₂ required : 0.054 g

Further explanation

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

4Fe(OH)⁺ (aq) + 4OH⁻ (aq) + O₂ (g)+ 2H₂O (l)⇒ 4Fe(OH)₃ (s)

To precipitate Fe²⁺(in Fe(OH⁺) to Fe³⁺(in Fe(OH)₃)

• mol Fe²⁺

$\tt 75\times 0.09=6.75~mlmol=6.75\times 10^{-3}mol$

mol ratio from equation = mol O₂ :mol Fe(OH) = 1 : 4

• mol O₂

$\tt \dfrac{1}{4}\times 6.75\times 10^{-3}=1.6875\times 10^{-3}$

• mass O₂

$\tt mass=mol\times MW\\\\mass=1.6875\times 10^{-3}\times 32=5.4\times 10^{-2}~g$