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.