Iron(II) sulfide reacts with hydrochloric acid according to the reaction:
FeS(s) + 2 HCl(aq) right arrow FeCl2(s) + H2S(g)
A reaction mixture initially contains 0.175 mol FeS and 0.481 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

Question

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:
FeS(s) + 2 HCl(aq) right arrow FeCl2(s) + H2S(g)
A reaction mixture initially contains 0.175 mol FeS and 0.481 mol HCl. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left?

aaronq · Answer

divide the moles of each by it's stoichiometric coefficient

$\dfrac{n_{FeS}}{1}=normalized\; moles$

determine which is the limiting reactant (the one with less normalized moles)

find how much is left by subtracting the moles of limiting reactant from the other reactant (the reactant with more moles)

anonymous · Answer

that's how I tried to do it which is wrong

aaronq · Answer

0.481/2 mol HCl = 0.2405

0.175/1 mol FeS = 0.175


0.2405-0.175=0.0655 

so you're left with 0.0655 moles of HCl