Iron(II) sulfide reacts with hydrochloric acid according to the
r...

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
remains?

Accepted Answer

Hi everyone today, we have a question giving us the reaction copper sulfide solid plus two hydrochloric acid. Acquis forms copper chloride, acquis plus hydrogen sulfide gasses And it tells us that initially there are 1.34 moles of copper sulfide and 4.29 moles of hydrochloric acid. And we need to calculate the moles of the excess reactant that remains after this reaction is complete. So first we have to identify our limiting re agent and that is going to be copper sulfide because it has the least number of moles. Now we can put that into an equation. So 1.34 moles of copper sulfide times two moles of hydrochloric acid because that's how many we have in our balanced equation. Over one mole of copper sulfide Equals 2.68 moles of hydrochloric acid. So now we need to take our 4.29 moles of hydrochloric acid minus r 2.68 moles. And that gives us 1.61 moles of hydrochloric acid. So our answer is D 1.61 moles. Thank you for watching. Bye.

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