Question

According to the following reaction, how many moles of Fe(OH)₂ can form from 175.0 ml of 0.227 m lioh solution?

assume that there is excess FeCl₂. FeCl₂(aq) 2 LiOH(aq) → Fe(OH)₂(s) 2 LiCl(aq)
a. 5.03 × 10⁻² moles
b. 6.49 × 10⁻² moles
c. 1.99 × 10⁻² moles
d. 2.52 × 10⁻² moles
e. 3.97 × 10⁻² moles

Answer 1

To find the number of moles of Fe(OH)₂ from 175.0 ml of 0.227 M LiOH solution, we use stoichiometry, leading to approximately 0.0199 moles or 1.99 × 10⁻² moles of Fe(OH)₂, which corresponds to option (c).

Step-by-step explanation:

To determine the number of moles of Fe(OH)₂ that can form from 175.0 ml of 0.227 M LiOH solution when there is excess FeCl₂, we can use stoichiometry. First, we need to convert the volume of LiOH solution to moles.

• Number of moles of LiOH = Volume (L) × Molarity (M)
• Number of moles of LiOH = 0.175 L × 0.227 mol/L = 0.039725 mol

According to the balanced equation FeCl₂(aq) + 2 LiOH(aq) → Fe(OH)₂(s) + 2 LiCl(aq), one mole of FeCl₂ reacts with two moles of LiOH to produce one mole of Fe(OH)₂. Hence, we have:

• Moles of Fe(OH)₂ = Moles of LiOH ÷ 2
• Moles of Fe(OH)₂ = 0.039725 mol ÷ 2 = 0.0198625 mol

So, the number of moles of Fe(OH)₂ that can form from 175.0 ml of 0.227 M LiOH solution is approximately 0.0199 moles, which can be rounded to 1.99 × 10⁻² moles, option (c).