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?

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Answer 1 · 2018-05-29T09:34:16

$37.15361$ g

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.