1. 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 $IO_3^-$.

A. 25.4 mL
B. 379 mL
C. 12.6 mL
D. 98.2 mL
E. 49.1 mL

2. How many mL of 1 M H₂SO₄ acid solution is required to neutralize 10 mL of 1 M NaOH?

A. 5 mL
B. 2.5 mL
C. 10 mL
D. 20 mL

(Note: The third question appears to contain incomplete and nonsensical information and is thus omitted.)

To calculate the volume of the KI solution required, we need to use the concept of stoichiometry and the balanced equation for the reaction.

First, let's determine the moles of KMnO4 using its concentration and volume:
Given: Concentration of KMnO4 = 0.216 N, Volume of KMnO4 solution = 34.1 mL
Using the formula: Moles = Concentration * Volume (in liters)
Converting the volume to liters: 34.1 mL = 0.0341 L
Calculating the moles of KMnO4: Moles of KMnO4 = 0.216 N * 0.0341 L = 0.0073756 moles
Next, using the balanced equation, we can determine the moles of KI required:
From the balanced equation: 3 moles of KMnO4 react with 2 moles of KI
Using the ratio: Moles of KI = (Moles of KMnO4 * 2) / 3 = (0.0073756 moles * 2) / 3 = 0.004917 moles
Finally, we can calculate the volume of the KI solution using its concentration:
Given: Concentration of KI = 0.150 N
Using the formula: Volume = Moles / Concentration (in liters)
Calculating the volume of KI solution: Volume = 0.004917 moles / 0.150 N = 0.03278 L
Converting the volume to milliliters: Volume = 0.03278 L * 1000 mL/L = 32.78 mL

Learn more about volume calculation in a chemical reaction here:

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