Question

55 mL of 1.0 M NH₃ is added to a flask containing the following solutions:

> 2.0 mL of 1.0 M FeCl₂ (in 2 M HCl) and

>2.0 mL of 2.0 M FeCl₃ (in 2 M HCl)

The ammonia first neutralizes the HCl in the mixed solutions and then reacts with the ferrous ion and ferric ion in the solution to make Fe₃O4.

If it takes 0.008 mol of NH₃ to neutralize the HCl, how many moles of NH3 are required to react with the Fe²⁺ and Fe³⁺ ions? How many moles NH3 will be leftover?

Answer 1

We know that Molarity = Moles/Volume(in L)

We are given the molarity and volume of NH₃, thus we can find the moles of NH₃.

Note that 55 mL = 55/1000 = 0.055 L

Moles of NH₃ = (1.0)*(0.055) = 0.055 moles

SImilarly, moles of Fe⁺²(FeCl₂) = (1.0)*(0.002) = 0.002 moles and moles of Fe⁺³ = (2.0)*(0.002) = 0.004 moles, that is total 0.006 moles.

We know that 0.008 moles of NH₃ are required to neutralize the HCl and moles of NH₃ required for reaction with Fe⁺² and Fe⁺³ ions in the mixed solution is 0.006.

Moles of NH₃ remaining = 0.055 - (0.008 + 0.006) = 0.041 moles