Question

According to the following reaction, how many moles of Fe(OH)2 can form from 175.0 mL of 0.227 M LiOH solution? Assume that there is excess FeCl2. FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)

Answer 1

Answer:

0.020 moles of $Fe(OH)_{2}$ can be formed

Explanation:

1. First determine the number of moles of LiOH.

Molarity is given by the following expression:

$M=\frac{molesofsolute}{Litersofsolution}$

Solving for moles of solute:

moles of solute = M * Liters of solution

Converting 175.0mL to L:

$175.0mL*\frac{1L}{1000mL}=0.175L$

Replacing values:

moles of solute = 0.227M*0.175L

moles of solute = 0.040

Therefore there are 0.040 moles of LiOH

2. Then write the balanced chemical reaction and use the stoichiometry of the reaction to calculate the number of moles of $Fe(OH)_{2}$ produced:

$FeCl_{2}(aq)+2LiOH(aq)=Fe(OH)_{2}(s)+2LiCl(aq)$

As the problem says that there are excess of $FeCl_{2}$, the limiting reagent is the LiOH.

$0.040molesLiOH*\frac{1molFe(OH)_{2}}{2molesLiOH}=0.020molesFe(OH)_{2}$ can be formed

Answer 2

According to the reaction, 0.0199 moles of Fe(OH)2 would be formed.

Stoichiometric reactions

From the equation of the reaction:

FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)

The mole ratio of LiOH and Fe(OH)2 is 2:1

Mole of 175 mL, 0.227 M LiOH = 0.175 x 0.227

                                               = 0.0397 mole

Equivalent mole of of Fe(OH)2 = 0.0397/2

                                                 = 0.0199 mole

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