Question

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:
FeS(s)+2HCl(aq)→FeCl2(s)+H2S(g)
A reaction mixture initially contains 0.208 mol FeS and 0.678 mol HCl.

Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

Answer 1

FeS(s) + 2 HCl(aq) ----------------------->   FeCl2(s) + H2S(g)

1 mol         2 mol                                        1 mol          1 mol

0.208 mol 0.678 mol               

here limiting reagent is FeS. HCl is excess reagent.

here we need only 0.416 mol of HCl. but we have 0.678 mol HCl

excess HCl = 0.678 - 0.416 = 0.262 mol

excess reactant left = 0.262 mol