Question

Iron (II) sulfide reacts with hydrochloric acid according to the reaction: FeS(s) + 2 HCl(aq)→ FeCl2(s) + H2S(g) A reaction mixture initially contains 0.223 mol FeS and 0.652 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

Answer 1

The amount ( in moles  of excess  reactant that is left is 0.206 moles

Explanation

FeS(s)  + 2HCl  (aq) →  FeCl2 (s)  + H2S (g)

• by use of mole ratio of FeS: HCl which is  1:2 this means that  0.223 mole  of FeS  reacted  completely with 0.223   x   2/1 =0.446  moles  0f FeCl2.

• HCl was in  excess because  0.446 moles of HCl reacted and initially  there was  0.652  moles.

• Therefore the  amount that was left

      = 0.652- 0.446  =0.206  moles

Answer 2

The amount of the excess reactant remaining after the reaction is complete is 0.206 mole

• We'll begin by determining the limiting reactant and the excess reactant. This can be obtained as follow:

FeS + 2HCl —> FeCl₂ + H₂S

From the balanced equation above,

1 mole of FeS reacted with 2 moles of HCl.

Therefore,

0.223 mole of FeS will react with = 2 × 0.223 = 0.446 moles of HCl

From the calculation made above, only 0.446 moles of HCl out of 0.652 mole reacted completely with 0.223 mole of FeS.

Therefore, FeS is the limiting reactant and HCl is the excess reactant.

• Finally, we shall determine the amount of the excess reactant remaining. This can be obtained as follow:

The excess reactant is HCl.

Amount given = 0.652 mole

Amount that reacted = 0.466 mole

Amount remaining =?

Amount remaining = (Amount given) – (Amount that reacted)

Amount remaining = 0.652 – 0.466

Amount remaining = 0.206 mole

Therefore, the amount of the excess reactant remaining is 0.206 mole

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