## Chemistry and Chemical Reactivity (9th Edition)

a) $0.625\ g$ b) $0.0252\ M$
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: $1.70/67.5= 0.0252\ M$