Question

How many moles of HCl would react with 37.1 mL of 0.138 M Sr(OH)2

Answer 1

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.