Question

1. the base sr(oh)₂ is used to neutralize 25.00 ml of the acid hcl according to the reaction below. the concentration of sr(oh)₂ is 0.113 m and the volume used to neutralize the acid was 16.50 ml sr(oh)₂. what is the concentration of hcl? reaction: 2hcl(aq) + sr(oh)₂(aq) → srcl₂(aq) + 2h₂o(l). (8 pts—work is 6 pts; answer is 2 pts) moles of hcl = moles of sr(oh)₂ × \\(\frac{2\text{ mol hcl}}{1\text{ mol sr(oh)₂}}\\) 2hcl(aq) + sr(oh)₂(aq) → srcl₂(aq) + 2h₂o hcl = 0.0018645 × 2 = 0.003729 mol molarity of hcl = \\(\frac{0.003729\text{ mol}}{0.02500\text{ l}}\\) = 0.14916 m = 0.149 m of hcl 9. what type of chemical reaction is the reaction in report question 8? (3 pts)

Explanation:

Step1: Calculate moles of \(Sr(OH)_2\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume).
\(n_{Sr(OH)_2}=0.113\space M\times0.01650\space L = 0.0018645\space mol\)

Step2: Calculate moles of \(HCl\)

From the reaction \(2HCl(aq)+Sr(OH)_2(aq)\to SrCl_2(aq) + 2H_2O(l)\), the mole ratio of \(HCl\) to \(Sr(OH)_2\) is \(2:1\).
\(n_{HCl}=n_{Sr(OH)_2}\times2=0.0018645\space mol\times2 = 0.003729\space mol\)

Step3: Calculate concentration of \(HCl\)

Use the formula \(C=\frac{n}{V}\) (where \(V = 25.00\space mL=0.02500\space L\)).
\(C_{HCl}=\frac{0.003729\space mol}{0.02500\space L}=0.14916\space M\approx0.149\space M\)

The reaction \(2HCl(aq)+Sr(OH)_2(aq)\to SrCl_2(aq)+2H_2O(l)\) is an acid - base neutralization reaction. In an acid - base neutralization reaction, an acid (\(HCl\)) reacts with a base (\(Sr(OH)_2\)) to form a salt (\(SrCl_2\)) and water (\(H_2O\)).

Answer:

The concentration of \(HCl\) is \(0.149\space M\)

For question 9: