What is the limiting reagent when 9.517 g of Fe is allowed to react with 19.34 g of water according to
1 answer:
You might be interested in
Answer:
The answer to your question is below
Explanation:
A.
[H₃O⁺] = 2 x 10⁻¹⁴ M
pH = ?
Formula
pH = - log [H₃O⁺]
Substitution
pH = - log [2 x 10⁻¹⁴]
Result
pH = 13.7
B.
[H₃O⁺] = ?
pH = 3.12
Formula
pH = - log [H₃O⁺]
Substitution
3.12 = - log [H₃O⁺]
Result
[H₃O⁺] = 7.59 M
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
- Oxidation reaction
Li⁰(s) → Li⁺(aq) + e⁻ (2)
- Reduction reaction
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
Learn more here:
I hope it helps you!
