We are asked to calculate the amount (in moles) of the excess reactant.

FeS(s) + 2 HCl(aq) → FeCl_{2}(s) + H_{2}S(g)

Calculate the mole FeCl_{2} based on both reactants:

$\mathbf{0}\mathbf{.}\mathbf{226}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{FeS}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}{\mathbf{FeCl}}_{\mathbf{2}}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{FeS}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{226}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{FeCl}}_{\mathbf{2}}\phantom{\rule{0ex}{0ex}}\mathbf{0}\mathbf{.}\mathbf{656}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}{\mathbf{FeCl}}_{\mathbf{2}}}{\mathbf{2}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{HCl}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{328}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}{\mathbf{FeCl}}_{\mathbf{2}}$

Since FeS produced less product, it is limiting.

Calculate HCl used.

$\mathbf{0}\mathbf{.}\mathbf{226}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{FeS}\mathbf{}}\mathbf{\times}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{\hspace{0.17em}}\mathbf{HCl}}{\mathbf{1}\mathbf{}\overline{)\mathbf{mol}\mathbf{}\mathbf{FeS}\mathbf{}}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{452}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{HCl}$

Iron(II) sulfide reacts with hydrochloric acid according to the reaction:

FeS(s) + 2 HCl(aq) → FeCl_{2}(s) + H_{2}S(g)

A reaction mixture initially contains 0.226 mol FeS and 0.656 mol HCl.

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

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