your error, I think, is you divided twice.
50ml of .153M is .00765mol KOH. That is three times as much as in your KOH used.
.00765*1/3*106.88=.273g
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 0.00383 mol Fe(NO3)3 remaining.
I don't know how to obtain the mass of Fe(OH)3, but here is my work:
(0.00255 mol KOH) x (1mol Fe(OH)3/3mol KOH) x (106.88 g/1mol Fe(OH)3
The answer key says 0.273 g Fe(OH)3 produced ( when I calculate the work is set up, this answer appeared when I did not divide it by 3, but there are 3 moles of KOH for every 1 mol of Fe(OH)3.)??
2 answers
Thank you!