Stoichiometry Question 2

\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?

\pu{0.0010 } M

\pu{0.010 } M

\pu{0.10 } M

\pu{1.0 } M

Approach

\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}