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If i perform this reaction by combining 25.0 grams of lioh with an excess of fe(no3)3, how much fe(oh)3 will i be able to make

Respuesta :

kirillzhulanov
3LiOH+Fe(NO3)3--›Fe(OH)3↓+3LiNO3
n(LiOH)=m/M=(25g)/(24g/mol)=1,0417mol
n(Fe(OH)3)=1/3n(LiOH)=0,3472mol
m(Fe(OH)3)=n*M=0,3472mol*107g/mol=37,15g

Mole is "an SI unit used to measure the amount of any substance".

What is LiOH?

LiOH is Lithium hydroxide.

What is Fe(OH)₃?

Fe(OH)₃ is iron (III) hydroxide.

What is Fe(NO₃)₃?

Fe(NO₃)₃ is iron (III) nitrate.

This is the above mentioned reaction,

3LiOH+Fe(NO₃)₃ →3LiNO₃+Fe(OH)₃

Number of moles of LiOH = 25/7+16+1

                                         = 1.0417 mol

Number of moles of Fe(OH)₃ = 1.0417/3 = 0.34723mol

         mass of Fe(OH)₃          =0.34723 x 56+3(16+1)

                                               = 0.34723 x 56+48+3

                                               = 37.15361g

Hence, 37.15361g of Fe(OH)₃ can be made from 25g of LiOH.

To learn more about moles here

https://brainly.com/question/8445690

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