Question

2.20 g of solid NaOH are added to 250 mL of a 0.10 M FeCl2 solution. Given that Ksp for Fe(OH)2 is 1.6 × 10-14, calculate the mass of Fe(OH)2 formed. State any assumptions you have made

Answer 1

Step-by-step explanation:

To calculate the mass of Fe(OH)2 formed, we first need to determine the limiting reagent. From the balanced equation:

2 NaOH + FeCl2 → Fe(OH)2 + 2 NaCl

We can see that the stoichiometric ratio between NaOH and Fe(OH)2 is 2:1. This means that we need twice as many moles of NaOH as moles of Fe(OH)2.

First, we calculate the moles of NaOH:

Mass of NaOH = 2.20 g

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol

Moles of NaOH = Mass of NaOH / Molar mass of NaOH

= 2.20 g / 39.99 g/mol

= 0.055 moles

Next, we calculate the moles of Fe(OH)2 that can potentially form.

Since the molarity of FeCl2 is 0.10 M and the volume is 250 mL (or 0.250 L), we have:

Moles of FeCl2 = Molarity of FeCl2 * Volume of FeCl2

= 0.10 M * 0.250 L

= 0.025 moles

However, the stoichiometric ratio between NaOH and Fe(OH)2 is 2:1. Therefore, we need to double the moles of FeCl2 to compare with the moles of NaOH.

Moles of Fe(OH)2 = 2 * Moles of FeCl2

= 2 * 0.025 moles

= 0.050 moles

Since the ratio between NaOH and Fe(OH)2 is 2:1, the moles of NaOH are in excess.

Next, we calculate the mass of Fe(OH)2 formed using the moles of Fe(OH)2 and its molar mass:

Molar mass of Fe(OH)2 = 55.85 g/mol + 16.00 g/mol + 2 * 1.01 g/mol

= 89.87 g/mol

Mass of Fe(OH)2 formed = Moles of Fe(OH)2 * Molar mass of Fe(OH)2

= 0.050 moles * 89.87 g/mol

= 4.49 g

Therefore, the mass of Fe(OH)2 formed is 4.49 g.

Answer 2

Assumptions:

1. The reaction goes to completion, and all the NaOH reacts with FeCl2 to form Fe(OH)2.

2. There are no other side reactions or interfering factors affecting the precipitation of Fe(OH)2.

3. The volume of the solution remains constant during the reaction.

4. The densities of the solutions are approximately equal to the density of water, so we do not need to account for volume changes.

Step-by-step explanation:

To calculate the mass of Fe(OH)2 formed, we first need to determine the amount of Fe(OH)2 that will precipitate from the reaction between NaOH and FeCl2. We can use the concept of solubility product constant (Ksp) to do this.

The balanced chemical equation for the reaction between NaOH and FeCl2 is:

2 NaOH + FeCl2 -> Fe(OH)2 + 2 NaCl

We can see that two moles of NaOH react with one mole of FeCl2 to produce one mole of Fe(OH)2.

Step 1: Calculate the moles of FeCl2 in the 250 mL solution.

Moles of FeCl2 = Volume (in liters) × Molarity

Moles of FeCl2 = 0.250 L × 0.10 mol/L = 0.025 mol

Step 2: Determine the limiting reactant.

Since the reaction ratio is 2:1 (2 moles of NaOH to 1 mole of FeCl2), we need to compare the moles of NaOH and FeCl2 to determine the limiting reactant.

Moles of NaOH = Mass ÷ Molar mass

Moles of NaOH = 2.20 g ÷ 40.00 g/mol (molar mass of NaOH) ≈ 0.055 mol

Comparing the moles of NaOH (0.055 mol) and FeCl2 (0.025 mol), we can see that FeCl2 is the limiting reactant because it is present in a lesser amount.

Step 3: Calculate the moles of Fe(OH)2 formed.

Since the molar ratio of Fe(OH)2 to FeCl2 is 1:1, the moles of Fe(OH)2 formed will be the same as the moles of FeCl2 reacted.

Moles of Fe(OH)2 formed = 0.025 mol

Step 4: Calculate the mass of Fe(OH)2 formed.

Mass of Fe(OH)2 formed = Moles of Fe(OH)2 × Molar mass

Mass of Fe(OH)2 formed = 0.025 mol × 89.86 g/mol (molar mass of Fe(OH)2) ≈ 2.25 g

Assumptions:

1. The reaction goes to completion, and all the NaOH reacts with FeCl2 to form Fe(OH)2.

2. There are no other side reactions or interfering factors affecting the precipitation of Fe(OH)2.

3. The volume of the solution remains constant during the reaction.

4. The densities of the solutions are approximately equal to the density of water, so we do not need to account for volume changes.