What mass of #Fe(OH)_3# is produced when 35 mL of 0.250 M #Fe(NO_3)_3# solution is mixed with 55 mL of a 0.180 M #KOH# solution?

The reaction between iron(III) nitrate, #"Fe"("NO"_3)_3#, and potassium hydroxide, #"KOH"#, will produce iron(III) hydroxide, #"Fe"("OH")_3#, an insoluble solid that precipitates out of solution, and aqueous potassium nitrate, #"KNO"_3#

#"Fe"("NO"_ 3)_ (3(aq)) + color(red)(3)"KOH"_ ((aq)) -> "Fe"("OH")_ (3(s)) darr + 3"KNO"_ (3(aq))#

The first thin to do here is use the molarities and volumes of the two solutions to figure out how many moles of each reactant you're mixing

#35 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.250 moles Fe"("NO"_3)_3)/(1 color(red)(cancel(color(black)("L solution")))) = "0.00875 moles Fe"("NO"_3)_3#

#55 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.180 moles KOH")/(1 color(red)(cancel(color(black)("L solution")))) = "0.00990 moles KOH"#

Now, do you have enough moles of potassium hydroxide to allow for all the moles of iron(III) nitrate to react? Use the #1:color(red)(3)# mole ratio that exists between the two reactants to figure this out

#0.00875 color(red)(cancel(color(black)("moles Fe"("NO"_3)_3))) *(color(red)(3)color(white)(.)"moles KOH")/(1color(red)(cancel(color(black)("mole Fe"("NO"_3)_3)))) = "0.02625 moles KOH"#

As you can see, you'd need #0.02625# moles of potassium hydroxide in order for all the moles of iron(III) hydroxide to react. Since you have fewer moles of potassium hydroxide than you'd need, potassium hydroxide will act as a limiting reagent.

This means that only

#0.00990 color(red)(cancel(color(black)("moles KOH"))) * ("1 mole Fe"("NO"_3)_3)/(color(red)(3)color(red)(cancel(color(black)("moles KOH")))) = "0.00330 moles Fe"("NO"_3)_3#

will take part in the reaction. Since iron(III) nitrate and iron(III) hydroxide are in a #1:1# mole ratio in the balanced chemical reaction, you can say that the reaction will produce #0.00330# moles of iron(III) hydroxide.

Use the compound's molar mass to convert this to grams

#0.00330 color(red)(cancel(color(black)("moles Fe"("OH")_3))) * "106.87 g"/(1color(red)(cancel(color(black)("mole Fe"("OH")_3)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.35 g")color(white)(a/a)|)))#

The answer is rounded to two sig figs.