In a reaction between XeO₃ and Mn²⁺ that occurred in 324 mL of solution, the gaseous Xe produced occupied a volume of 595 mL at STP. What is the molarity of the MnO₄^- solution resulting from the reaction?
9 H₂O + 6 Mn²⁺ + 5 XeO₃ → 6 MnO₄^- + 5 Xe + 18 H⁺

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 5 moles of XeO3 react with 6 moles of Mn2+ to produce 6 moles of MnO4-. The volume of Xe gas produced is 595 mL at STP, which is equivalent to 0.595 L. Using the ideal gas law, we can calculate the number of moles of Xe produced:PV = nRTn = PV/RT = (1 atm)(0.595 L)/(0.0821 L·atm/mol·K)(273 K) = 0.025 molSince 5 moles of XeO3 produce 6 moles of MnO4-, 0.025 mol of XeO3 will produce 0.03 mol of MnO4-. The molarity of the MnO4- solution is therefore:M = n/V = 0.03 mol / 0.324 L = 0.0926 MThe closest answer choice is A) 0.0983 M.

In a Reaction Between XeO3 and Mn2+ That Occurred in 324

In a Reaction Between XeO₃ and Mn²⁺ That Occurred in 324