Optimistic Chemistry Challenge: Neutralizing Acid Spills!

How much barium hydroxide is needed to neutralize a spill of 325 grams HCl?

A. 450 grams

B. 600 grams

C. 762 grams

D. 900 grams

Answer:

To neutralize 325 grams of HCl, 762 grams of barium hydroxide is needed.

The subject of this chemistry challenge involves the neutralization reaction between HCl (hydrochloric acid) and Ba(OH)2 (barium hydroxide), showcasing the fascinating world of chemical reactions. In this scenario, we aim to determine the minimum amount of barium hydroxide required to completely neutralize a spill of 325 grams of HCl.

From the balanced chemical equation provided, it is evident that one molecule of barium hydroxide reacts with two molecules of hydrochloric acid to form barium chloride and water. This information allows us to establish the stoichiometry relationship required for the calculation.

Given that the molar mass of HCl is approximately 36.5 grams/mol, the spill of 325 grams corresponds to around 8.9 moles of HCl. According to the stoichiometry of the reaction, one mole of barium hydroxide is needed to neutralize two moles of HCl, resulting in a total of 4.45 moles of Ba(OH)2 required.

Considering the molar mass of Ba(OH)2 is approximately 171.34 grams/mol, the final amount of barium hydroxide needed can be calculated as 4.45 moles multiplied by 171.34 grams/mol, resulting in approximately 762 grams of Ba(OH)2. Therefore, the correct answer is option C: 762 grams.

Chemistry challenges like this not only enhance our understanding of chemical reactions but also demonstrate the practical applications of chemical concepts in real-world scenarios. Keep exploring the fascinating realm of chemistry!