Question

A precipitate of iron (III) hydroxide, Fe(OH)3, forms when solutions of iron (111) nitrate, Fe(NO3)3, and

potassium hydroxide, KOH, are combined:
Fe(NO3)2 (aq) + 3 KOH(aq) + 3 KNO3 (aq) + Fe(OH)3(s)


Predict the mass of iron (III) hydroxide is expected to form when 130.0 mL of 0.40 mol/L of potassium
hydroxide is added to excess iron (III) nitrate solution.

Make sure to show all your work including molar mass calculations.

Answer 1

Approximately 1.849 grams of iron (III) hydroxide is expected to be formed when 130.0 mL of 0.40 mol/L of potassium hydroxide is mixed with excess iron (III) nitrate solution.

Step-by-step explanation:

To predict the mass of iron (III) hydroxide expected to form, we must first calculate the number of moles of KOH added. Since concentration is given in mol/L and we have a volume in mL, we convert the volume to liters:

130.0 mL = 0.130 L

Then, using the molarity:
0.40 mol/L × 0.130 L = 0.052 mol KOH

According to the balanced chemical equation for this reaction (which should be Fe(NO3)3+ 3KOH → 3KNO3 + Fe(OH)3), one mole of Fe(OH)3 forms per three moles of KOH. Hence, only 0.052/3 = 0.0173 mol of Fe(OH)3 can be produced.

To find the mass of Fe(OH)3, calculate the molar mass:
55.845 (Fe) + 3 × 15.999 (O) + 3 × 1.008 (H) = 106.867 g/mol

Finally, multiply the number of moles of Fe(OH)3 by its molar mass:

0.0173 mol × 106.867 g/mol = 1.849 g of Fe(OH)3

The mass of iron (III) hydroxide expected to form when 130.0 mL of 0.40 mol/L potassium hydroxide is added to excess iron (III) nitrate solution is approximately 1.849 g.