Question

How many moles of fe2+ions are formed when excess iron is treated with 500 ml of 0.4m hcl under inert atmosphere ? assume no change in volume?

Answer 1

2HCl + Fe ---> Fe2+ + H2 + 2Cl- 
But you can rewrite it as 2H+ + Fe ---> Fe2+ + H2 if you want. The Cl- ions are just spectators (they don't take place in the reaction itself so you can remove them since HCl breaks up into H+ and Cl- because it is one of the 7 strong acids) 

So we can use the first equation for this question to make it simpler. 
2HCl + Fe ---> Fe2+ + H2 + Cl- 

Now, we want to start with moles of HCl and we are conveniently given its molarity and volume. 
Molarity * Volume (in Liters) = moles 
4M * (.05L) = 0.2 mol HCl 

Now it becomes simple stoichiometry! 
0.2 mol HCL * [1 mol Fe2+ / 2 mol HCl] 
= 0.2/2 
= 0.1 mol Fe 2+ ions

Answer 2

HELLO THERE!

The reaction is:

Fe + 2HCl ----> FeCl₂ + H₂

Molality of the HCl solution = 0.4 m

So, 0.4 moles of HCl is present in 1000 grams of the solvent.

According to my knowledge, some data in your question is missing, if the molality is given. So, your question must be 500 mL of 0.4 M solution.

If so, then 0.4 moles of HCl is present in 1000 mL of it.

So, HCl present in 500 mL of it =

$\frac{0.4}{1000}\times500 = 0.2 mol$

According to the equation:

Fe + 2HCl ----> FeCl₂ + H₂

2 moles of HCl produce 1 mole of Iron chloride.

So, 0.2 moles of HCl produce =

$\frac{1}{2}\times0.2 = 0.1$

= 0.1 moles of Iron chloride.

Now, 1 mole iron chloride contains 1 mole Fe²⁺ ion.

So, 0.1 moles of iron chloride will contain 0.1 moles of Fe²⁺ ion.

THIS IS YOUR ANSWER.

HOPE IT HELPS!