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3 years ago

Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)

Chemistry

1 answer:

Brilliant_brown [7]3 years ago

3 0

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 two reactions, 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 losing one electron) in the lithium acetate (reaction 2) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by gaining two electrons) (reaction 3).

We must remember that the reducing agent is the one that will be oxidized by reducing another element and that the oxidizing agent is the one that will be reduced by oxidizing another species.

In reaction (1), the reducing agent is Li (it is oxidizing to Li⁺), and the oxidizing agent is Fe(CH₃COO)₂ (it is reducing to Fe⁰).

Therefore, the reducing agent in reaction (1) is lithium (Li).

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3 years ago

The equation for another reaction used in industry isCO(g) + H₂O(g) <img src="https://tex.z-dn.net/?f=%5Crightleftharpoons" id="

Sloan [31]

Answer:

(i) CO = 0.4 mol; H₂O = 1.6 mol; Kc = 4

(ii) CO = 0.67 mol; H₂O = 0.67 mol; CO₂ = 1.33 mol; H₂ = 1.33 mol

Explanation:

(i) For the equation given let's make a table of the concentrations for equilibrium (the volume is constant, so, we can do it with moles number)

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 3.2 mol 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x 3.2-x x x Equilibrium

In the equilibrum, the moles number of hydrogen and carbon dioxide are 1.6 mol, so x = 1.6 mol

The amounts of CO and H₂O are:

CO = 2.0 - 1.6 = 0.4 mol

H₂O = 3.2 - 1.6 = 1.6 mol

The constant of the equilibrium is the multiplications of the concentrations of products divided by the multiplication of the concentration of the reactants (all the concentrations elevated to the coefficient). So:

Kc = (1.6x1.6)/(0.4x1.6)

Kc = 1.6/0.4

Kc = 4

(ii) Kc must remais constant (it only changes with the temperature), so let's construct a new table of equilibrium:

CO(g) + H₂O(g) ⇄ H₂(g) + CO₂(g)

2.0 mol 2.0 mol 0 0 Initial

-x -x +x +x Reacts (stoichiometry is 1: 1: 1: 1)

2.0-x 2.0-x x x Equilibrium

Kc = (x*x)/((2.0-x)*(2.0-x))

4 = x²/(4 - 4x + x²)

16 - 16x + 4x² = x²

3x² - 16x + 16 = 0

Using Baskhara's equation:

Δ =(-16)² - 4x3x16

Δ = 256 - 192

Δ = 64

x = (-(-16) +/- √64)/(2*3)

x' = (16 + 8)/6 = 4

x'' = (16 - 8)/6 = 1.33

x must be small than 2.0, so x = 1.33 mol, which is the amount of hydrogen and carbon dioxide at equilibrium. The both reactants has 2.0 - 1.33 = 0.67 mol at equilibrium.

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3 years ago

Arrange the following elements in order of increasing (from smallest to largest) atomic size: H, Br, Cl, O. Explain the reasonin

juin [17]

Explanation:

Atomic size is defined as the distance between the nucleus and valence shell of an atom. Valence shell is the outermost shell of an atom.

When we move across a period in a periodic table then there will occur a decrease in atomic size of the atoms. But when we move across a group in a periodic table then there will occur an increase in atomic size of the atom.

Hydrogen is a period 1 element, oxygen is a period 2 element, chlorine is a period 3 element and bromine is a period 4 element.

Therefore, given elements are arranged in order of increasing atomic size as follows.

H < O < Cl < Br

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