This is a limiting reagent (LR) problem. You know that because an amount is given for both reactants.
3KOH + Fe(NO3)3 ==> Fe(OH)3 + 3KNO3
mols KOH = M x L = ?
mols Fe(NO3)3 = M x L = ?
Using the coefficients in the balanced equation, convert mols KOH to mols Fe(OH)3.
Do the same to convert mols Fe(NO3)3 to mols Fe(OH)3.
It is likely that these two values for Fe(OH)3 will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that smaller number is the LR. (The other (excess) reagent I will call the ER.)
To fine grams Fe(OH)3 it is grams = mols Fe(OH)3 x molar mass Fe(OH)3.
Now to find mols of ER. That is convert mols of LR used (all of it) to mols ER used. Then mols ER remaining = initial mols ER - mols ER used = ?
Calculate the mass of Fe(OH)3 s) produced by mixing 50.0 mL of 0.153 M KOH (aq) and 25.0 mL of 0.255 M Fe(NO3)3 (aq), and the number of moles of the excess reactant remaining in solution.
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