In a reaction between XeO₃(aq)and Mn²⁺(aq)that occurred in 324 mL of aqueous solution,the gaseous Xe produced occupied a volume of 595 mL at STP.What is the molarity of the MnO₄^-(aq)solution resulting from the reaction? 9 H₂O(l)+ 6 Mn²⁺(aq)+ 5 XeO₃(aq)→ 6 MnO₄^-(aq)+ 5 Xe(g)+ 18 H⁺(aq) A)0.0983 M B)0.0683 M C)0.00292 M D)0.0820 M E)0.00203 M

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 5 moles of XeO₃ react to produce 6 moles of MnO₄^-. The volume of Xe gas produced at STP (595 mL) corresponds to 595 mL / 22.4 L/mol = 0.0266 moles of Xe. Since the stoichiometry is 5 moles of XeO₃ to 6 moles of MnO₄^-, the moles of MnO₄^- produced are (0.0266 moles Xe) * (6/5) = 0.0319 moles. The molarity of MnO₄^- is then 0.0319 moles / 0.324 L = 0.0983 M.

In a Reaction Between XeO3(aq)and Mn2+(aq)that Occurred in 324 ML

In a Reaction Between XeO₃(aq)and Mn²⁺(aq)that Occurred in 324 ML