3LiOH+Fe(No3)3 →3LiNO3+Fe(OH)3 If I perform this reaction by combining 25g of LiOH with an excess of Fe(NO3)3, how much Fe(OH)3 will I be able to make?

I just need to to know the answer and an explanation for how to do it.

1 Answer
May 29, 2018

37.15361g

Explanation:

If "Fe"("NO"_3)_3 is excess, then "LiOH" is limiting.

We have to find the number of moles of limiting reagent,

no. of moles of "LiOH"=25/(7+16+1)=1.0417mol

no. of moles of "Fe"("OH")_3 to be made=1.0417/3=0.34723mol

mass of "Fe"("OH")_3 to be made=0.34723*(56+3(16+1))=37.15361g

You can theoretically produce 37.15361g out iron(III) hydroxide from the reagents you provided.