Chemical Reaction: HCl and Ca(OH)2

How can we determine if 2.8 mol of Ca(OH)2 is sufficient to neutralize 6.3 mol of HCl?

Can you calculate the amount of Ca(OH)2 needed to react with the spilled HCl?

Answer:

The question involves a chemical reaction between HCl and Ca(OH)2, asking whether 2.8 mol of Ca(OH)2 is sufficient to neutralize 6.3 mol of HCl, using stoichiometry from the balanced chemical equation.

The question pertains to a chemical reaction where 6.3 mol of HCl (hydrochloric acid) is spilled and then 2.8 mol of Ca(OH)2 (calcium hydroxide) is used to neutralize it. To solve this, we would need to look at the stoichiometry of the reaction between HCl and Ca(OH)2.

Based on a balanced chemical reaction, 2 mol of HCl reacts with 1 mol of Ca(OH)2. Therefore, theoretically, 2 mol of HCl would be completely neutralized by 1 mol of Ca(OH)2. Applying this ratio to the given moles:

2 mol HCl : 1 mol Ca(OH)2

6.3 mol HCl : x mol Ca(OH)2

To find the value of 'x,' which represents the moles of Ca(OH)2 needed to react with 6.3 mol of HCl, one can set up a proportion. After calculating, we can see whether 2.8 mol of Ca(OH)2 is sufficient to neutralize the spilled 6.3 mol of HCl.