Answer
a) $0.625\ g$
b) $0.0252\ M$
Work Step by Step
Number of moles of $FeCl_3$:
$25.0\ mL\times 0.234\ M=5.85\ mmol$
Number of moles of $NaOH$:
$42.5\ mL\times 0.453\ M=19.25\ mmol$
Ratio: $3.29\gt3$ $FeCl_3$ it's the limiting reactant.
a) Mass of iron hydroxide
$5.85\ mmol\times 106.87\ g/mol=0.625\ g$
b) Number of moles of excess:
$19.25-3\times5.85=1.70\ mmol$
Concentration: $1.70/67.5= 0.0252\ M$
