In a precipitation reaction between FeCl2(aq) and LiOH(aq), 11.6 mL of 0.210 M FeCl2(aq) completly reacted with 35.1 mL of LiOH(aq). What was the molarity of LiOH(aq)? FeCl2(aq) + 2 LiOH(aq) → Fe(OH)2(s) + 2 LiCl(aq)

The molarity of LiOH is 0.139 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

    .....(1)

Molarity of iron (II) chloride solution = 0.210 M

Volume of solution = 11.6 mL = 0.0116 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

For the given chemical reaction:

By Stoichiometry of the reaction:

1 mole of iron (II) chloride reacts with 2 moles of LiOH

So, moles of iron (II) chloride will react with = of LiOH

Calculating the molarity of LiOH by using equation 1:

Moles of LiOH =

Volume of LiOH = 35.1 mL = 0.0351 L

Putting values in equation 1, we get:

Hence, the molarity of LiOH is 0.139 M

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