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?

Question

Chemistry

Frogger

8 April, 01:03

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?

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Answers (1)

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Deangelo Ramirez · 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