Answer :
Answer is: 0.102 moles of HCl would react.
Balanced chemical reaction:
2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).
V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.
V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.
c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.
n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).
n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.
n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.
From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.
n(HCl) = 2 · n(Sr(OH)₂).
n(HCl) = 2 · 0.0051 mol.
n(HCl) = 0.0102 mol; amount of the hydrochloric acid.
Final answer:
The question involves stoichiometry in a chemical reaction between hydrochloric acid (HCl) and strontium hydroxide (Sr(OH)2). The provided volume and molarity of Sr(OH)2 allow us to calculate the needed moles of HCl, which revolves around a 2:1 stoichiometric ratio. Therefore, to react with 37.1 mL of 0.138 M Sr(OH)2, we need 0.0102 moles of HCl.
Explanation:
The subject matter involves a chemical reaction between hydrochloric acid (HCl) and strontium hydroxide (Sr(OH)2). First, we need to identify the stoichiometric ratio between HCl and Sr(OH)2. The chemical equation for their reaction is 2HCl + Sr(OH)2 → SrCl2 + 2H2O, indicating a 2:1 ratio between HCl and Sr(OH)2.
Next, calculate the moles of Sr(OH)2 using the given volume and molarity: 37.1 mL Sr(OH)2 * (1 L/1000 mL) * 0.138 mol/L = 0.0051 mol of Sr(OH)2. Since the ratio of HCl to Sr(OH)2 is 2:1, this means we would need 2 * 0.0051 = 0.0102 moles of HCl for the reaction.
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