Question

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.

Answer 1

`37.15361`g

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)=1.0417`mol

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

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

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