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shepuryov [24]
3 years ago
13

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

Chemistry
1 answer:
Solnce55 [7]3 years ago
5 0

mass O₂ required : 0.054 g

<h3>Further explanation </h3>

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

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