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29 May, 23:18

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

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  1. 30 May, 02:04
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    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
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