What volume of a 0.150 N KI solution is required to react in basic solution with 34.1 mL of a 0.216 N solution of KMnO₄? The products in the reaction include MnO₂ and IO₃⁻.

A. 25.4 mL
B. 37.9 mL
C. 12.6 mL
D. 98.2 mL
E. 49.1 mL

To complete the chemical reaction, 49.1 ml of 0.150 N KI solution is required to react with 34.1 ml of 0.216 N KMnO4 solution in the given stoichiometric problem.

This question pertains to a stoichiometric problem in chemistry involving a reaction between Potassium Iodide (KI) and Potassium Permanganate (KMnO4). In this redox reaction, the key point is that 3 moles of KI react with 1 mole of KMnO4. Thus, the molar ratio of KMnO4 to KI in the solution is 1:3.To solve the problem, we can use the formula: [tex]Volume of KI (V1) * Normality of KI (N1) = Volume of KMnO4 (V2) *[/tex] Normality of KMnO4 (N2).Plugging in the given values: V1 * 0.150 N = 34.1 ml * 0.216 N .Solving for V1, we get [tex]V1 = (34.1 ml * 0.216 N) / 0.150 N = 49.1[/tex]ml.Therefore, option (e) is the correct answer. To complete the reaction, 49.1 ml of 0.150 N KI solution is needed to react with 34.1 ml of 0.216 N KMnO4 solution.

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What volume of a 0.150 N KI solution is required to react in basic solution with 34.1 mL of a 0.216 N solution of KMnO₄? The products in the reaction include MnO₂ and IO₃⁻.A. 25.4 mL B. 37.9 mL C. 12.6 mL D. 98.2 mL E. 49.1 mL

Answer :

Other Questions

What volume of a 0.150 N KI solution is required to react in basic solution with 34.1 mL of a 0.216 N solution of KMnO₄? The products in the reaction include MnO₂ and IO₃⁻.

A. 25.4 mL
B. 37.9 mL
C. 12.6 mL
D. 98.2 mL
E. 49.1 mL