Question

When 50.0 mL of a 1.00 M solution of Fe(NO3)3 are mixed with 50.0mL of a 1.00 M solution of NaOH, a precipitate forms. What Ions remain after the reaction is complete?
Fe(NO3)3 (aq) + 3 NaOH (aq) -> Fe(OH)3 (s) + 3 NaNO3 (aq)

A. Fe3+, OH-, Na+, and NO3-
B. Fe3+ and OH-
C. Na+ and NO3-
D. Fe3+, Na+, and OH-

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NaOH = 1 x 50 x 10-3 = 50 x 10-3

moles of Fe(NO3)3 = 1 x 50 x 10-3 = 50 x 10-3

now consider the given reaction

Fe(NO3)3 (aq) + 3 NaOH (aq) -> Fe(OH)3 (s) + 3 NaNO3 (aq)

we can see that

moles of NaoH required = 3 x moles of Fe(NO3)3

moles of NaOH required = 3 x 50 x 10-3 = 150 x 10-3

but

only 50 x 10-3 moles of NaoH is present

so

NaOH is the limiting reagent

and

Fe(NO3)3 is the excess reactant

So

not all teh Fe+2 is precipitated

some Fe(N03)3 is left

So Fe+3 and N03- are present

from the reaction NaN03 is formed

So Na+ is also present

OH- is already present in the soluton as water

so

Fe+3 , OH- , Na+ , N03- are all present

so

the answer is option A