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Question: What mass of \(Fe{{(OH)}_{3}}\) is produced when 35 mL of 0.250 M \(Fe{{(N{{O}_{3}})}_{3}}\) solutio...

What mass of Fe(OH)3Fe{{(OH)}_{3}} is produced when 35 mL of 0.250 M Fe(NO3)3Fe{{(N{{O}_{3}})}_{3}} solution is mixed with 55 mL of a 0.180 M KOH solution?

Explanation

Solution

Before answering this question, we should know when iron (III) nitrate Fe(NO3)3Fe{{(N{{O}_{3}})}_{3}} reacts with potassium hydroxide KOHKOH, it will produce iron (III) hydroxide Fe(OH)3Fe{{(OH)}_{3}} and also a solid that is insoluble and precipitates out of solution and aqueous potassium nitrate KNO3KN{{O}_{3}}.

Complete answer:
Fe(NO3)3(aq)+3KOH(aq)Fe(OH)3(s)+3KNO3(aq)Fe{{(N{{O}_{3}})}_{{{3}_{(aq)}}}}+3KO{{H}_{(aq)}}\to Fe{{(OH)}_{{{3}_{(s)}}}}\downarrow +\,3KN{{O}_{{{3}_{(aq)}}}}
Now, we have to see the molarities and volumes of the two solutions in order to find out the number of moles of each reactant we are using.
35mL×1L103mL×0.250molesFe(NO3)31Lsolution=0.00875molesFe(NO3)335mL\times \dfrac{1L}{{{10}^{3}}mL}\times \dfrac{0.250\,moles\,Fe{{(N{{O}_{3}})}_{3}}}{1L\,solution}=0.00875\,moles\,Fe{{(N{{O}_{3}})}_{3}}
55mL×1L103mL×0.180molesKOH1Lsolution=0.00990molesKOH55mL\times \dfrac{1L}{{{10}^{3}}mL}\times \dfrac{0.180\,moles\,KOH}{1L\,solution}=0.00990\,moles\,KOH
Now, we have to see the number of moles of potassium hydroxide that are present, are they enough for all the moles of iron (III) nitrate to react?
We will use a 1:3 mole ratio that is between the two reactants to find out-
0.00875molesFe(NO3)3×3molesKOH1moleFe(NO3)3=0.02625molesKOH0.00875\,moles\,Fe{{(N{{O}_{3}})}_{3}}\times \dfrac{3\,moles\,KOH}{1\,mole\,Fe{{(N{{O}_{3}})}_{3}}}=0.02625\,moles\,KOH
Now, we need 0.026525 moles of potassium hydroxide so that all the moles of iron (III) hydroxide can react. The moles of potassium hydroxide is less than what we need so potassium hydroxide will work as a limiting reagent.
It means,
0.00990 moles KOH ×1molesFe(NO3)33molesKOH\times \dfrac{1\,moles\,Fe{{(N{{O}_{3}})}_{3}}}{3\,moles\,KOH}= 0.00330 moles Fe(NO3)3Fe{{(N{{O}_{3}})}_{3}}
It will be active in the reaction because iron (III) nitrate and iron (III) hydroxide are in a mole ratio 1:1 in the chemical equation which is balanced, this balanced equation will produce 0.00330 moles of iron (III) hydroxide.
Now, we will convert it into grams using the compound's molar mass.
0.00330molesFe(OH)3×106.87g1moleFe(OH)30.00330\,moles\,Fe{{(OH)}_{3}}\times \dfrac{106.87g}{1\,mole\,Fe{{(OH)}_{3}}}= 0.35 grams

Note:
Uses of potassium hydroxide (KOH)-
It helps to prepare the salts.
It helps in the neutralization of acids.
It helps in the production of soaps.
It also helps to manufacture batteries and fuel cells.