High School

If 22.3 mL of 2.60 M acetic acid reacts with 39.4 mL of 0.900 M radium hydroxide, which of the following best describes the resulting solution?

A. Limited
B. Basic
C. Acidic
D. Excess

Answer :

Final answer:

The reaction between acetic acid and radium hydroxide yields one mole of water for every one mole of each reactant. As the moles of radium hydroxide are less, it's the limiting reactant. Therefore, the resulting solution is described as having excess acetic acid.

Explanation:

The question is about a reaction between acetic acid (HC₂H3O2) and radium hydroxide. In a reaction between an acid and a base, the result depends on the amounts of each reactant used. These solutions are reacting in a neutralization reaction, where an acid reacts with a base to produce a salt and water.

First, we must understand that concentration (M) is defined as mol/L. So, if we multiply the volume of each solution (in L) by their concentration (in mol/L), we could find moles of each reactant. For acetic acid, it would be 22.3 mL * 2.60 M = 0.058 mol and for radium hydroxide, it would be 39.4 mL * 0.900 M = 0.035 mol.

However, the reaction between acetic acid and radium hydroxide yields one mole of water for every one mole of each reactant. This means, radium hydroxide is the limiting reactant, while acetic acid is in excess.

Therefore, the resulting solution would correspond to Option D: Excess. It's called so because there is an excess of acetic acid in the solution.

Learn more about Neutralization Reaction here:

https://brainly.com/question/11403609

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