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:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
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).
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Noble Gas. Metals have 1 or 2 Valence Electrons. Halogens have 7 Valence Electrons. Semi-Metals can have different amounts.
The concentration of diluted solution is 0.756M.
From the question given above, the following data were obtained:
Volume of stock solution (V1) = 18.9 mL
Molarity of stock solution (M1) = 10 M
Volume of diluted solution (V2) = 250 mL
Molarity of diluted solution (M2) =?
We can obtain the molarity of the diluted solution by using the dilution formula as shown follow:
M1V1 = M2V2
10 × 18.9 = M2 ×250
189 = M2 × 250
Divide both side by 100
M2 = 189 / 250
M2 = 0.756 M
Therefore, the molarity of the diluted solution is 0.756 M.
Thus the concluded that concentration of the dilute acid is 0.756 M.
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An exemple of aluminium in its ore state is haemetite