$$ \ce{NaOCl(aq) + 2HCl(aq) -> Cl2(g) + NaCl(aq) + H2O(l)} $$

Suppose an excess of \(\pu{6 }M  \ce{HCL(aq)}\) solution is added to \(100\). \(\pu{mL}\) of \(\ce{NaOCl(aq)}\) of unknown concentration. If the reaction goes to completion and \(\pu{0.010 mol}\) of \(\ce{Cl2}\) is produced, then what was the molarity of the \(\ce{NaOCl(aq)}\) solution?

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$$ \pu{0.010 mol }\ce{Cl2} \cdot \frac{\pu{1 mol }\ce{NaOCl}}{\pu{1 mol }\ce{Cl2}} \cdot \frac{1}{\pu{0.10 L}}$$ $$ =\frac{\pu{0.10 mol }\ce{NaOCl}}{\pu{1 L}} $$ $$ = \boxed{\pu{0.10 }M} $$