Answer
0.206 mol $HCl$
Work Step by Step
1. Identify the limiting reactant.
Find the amount of product if each reactant is completely consumed.
$$ 0.223 \space mole \space FeS \times \frac{ 1 \space mole \ H_2S }{ 1 \space mole \space FeS } = 0.223 \space mole \space H_2S $$
$$ 0.652 \space mole \space HCl \times \frac{ 1 \space mole \ H_2S }{ 2 \space moles \space HCl } = 0.326 \space mole \space H_2S $$
Since the reaction of $ FeS $ produces less $ H_2S $ for these quantities, it is the limiting reactant.
2. Find the amount of $ HCl $ consumed:
$$ 0.223 \space moles \space FeS \times \frac{ 2 \space mol \space HCl }{ 1 \space mol \space FeS } = 0.446 \space mol \space HCl $$
3. Subtract that amount from the amount in the reaction mixture.
$$Excess = 0.652 - 0.446 = 0.206 \space mol \space HCl $$
