Question

What mass of Fe(OH)3 will be obtained when 100. mL of 0.240 M FeCl3 is mixed with 200. mL of 0.182 M NaOH?

Answer 1

Answer:

648.68 mg

Explanation:

The reaction that takes place is:

• FeCl₃ + 3NaOH → Fe(OH)₃ + 3NaCl

First we calculate how many moles of each reactant were added, using the given volumes and concentrations:

• FeCl₃ ⇒ 100 mL * 0.240 M = 24 mmol FeCl₃

• NaOH ⇒ 100 mL * 0.182 M = 18.2 mmol NaOH

24 mmol of FeCl₃ would react completely with (24 * 3) 72 mmol of NaOH. There are not as many NaOH mmoles, so NaOH is the limiting reactant.

Now we calculate how many moles of Fe(OH)₃ are formed, using the moles of the limiting reactant:

• 18.2 mmol NaOH * $\frac{1mmolFe(OH)_3}{3mmolNaOH}$ = 6.07 mmol Fe(OH)₃

Finally we convert 6.07 mmol Fe(OH)₃ to grams, using its molar mass:

• 6.07 mmol Fe(OH)₃ * 106.867 mg/mmol = 648.68 mg