I agree with 3.83 mmols ferric but the problem asks for mols and not mmols.
I don't agree with 0.242 g Fe(OH)3 formed
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.
I got 0.242 g Fe(OH)3 and 3.83 mmol Fe3+ remaining.
2 answers
for each of the following reactions 3.00 moles of each reactant is present initially. Determine the limiting reactant and calculate the moles of product in parentheses that would form
Fe2O3 +3H2=2Fe3H2O
Fe2O3 +3H2=2Fe3H2O