You must 25.0 mL of 0.234 M FeCl 3 with 42.5 mL of0.453 M NaOH. (a) What mass of

Question

You must 25.0 mL of 0.234 M FeCl3 with 42.5 mL of0.453 M NaOH.

     (a) What mass ofFe(OH)3 (in grams) will precipitate from this reactionmixture?

     (b) One of the reactants(FeCl3 or NaOH) is present in a stoichiometricexcess.

     (c)     Whatis the molarity, or molar concentration, of the excess reactantremaining in

          solution after Fe(OH)3 has been precipitated?

Explanation / Answer

Moles FeCl3 = 25 x 0.234 / 1000 = 0.00585 => moles Fe3+ Moles NaOH = 42.5 x 0.453 / 1000 = 0.0193 => moles OH- Fe3+ + 3OH- >> Fe(OH)3 the ratio is 1 : 3 0.00585 x 3 = 0.0176 moles OH- needed FeCl3 is the limiting reactant.We get 0.00585 moles Fe(OH)3 MM = 106.87 g/mol grams Fe(OH)3 = 106.87 g/mol x 0.00585 = 0.625 g 0.0193 - 0.0176 = 0.00170 moles OH- in excess Molar mass NaOH = 39.997 g/mol Grams NaOH in excess =39.997 g/mol x 0.00170 mol =0.0680 g OR you can write it this way if you understand it easier mole= M*V, so for FeCl3 =0.234*25*10^-3= 5.85*10^-3 NaOH =0.453*42.5*10^-3 =19.25*10^-3 FeCl3 + 3NaOH -> Fe(OH)3 + 3NaCl so 1mole FeCl3 reacts with 3mole NaOH 5.85*10^-3 mole react with x Thus x =3*5.85*10^-3=17.55*10^-3 mole NaOH but we have 19.25*10^-3mole NaOH, therefore NaOH is in excess Therefore the amount of Fe(OH)3 will be calculated according to thelimiting reageant FeCl3. so from the reaction 1 mole FeCl3 gives 1 mole Fe(OH)3 or better yet, since the molecular weight of Fe(OH)3 is 106.85 1 mole FeCl3 gives 106.85 g Fe(OH)3 5.85*10^-3 mole give x g x= 106.85*5.85*10^-3= 0.625 g We calculated that 17.55*10^-3 mole NaOH reacted. So we have remaining (19.25-17.55) *10^-3= 1.7*10^-3 mole NaOH These are in a total volume of 25.0+42.5=67.5 ml= 67.5*10^-3L. So the concentration of the excess NaOH is C=mole/V= 1.7*10^-3/67.5*10^-3= 0.025 M

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