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hodyreva [135]
3 years ago
9

Determine the oxidation number of sulfur in S203 ( with steps)

Chemistry
1 answer:
sdas [7]3 years ago
8 0

Answer:

The oxidation number of sulphur in is +2

Explanation:

<em>S</em><em>2</em><em>O</em><em>3</em><em> </em><em>2</em><em>-</em>

<em>2</em><em> </em><em>(</em><em>S</em><em>)</em><em> </em><em>+</em><em> </em><em>3</em><em>(</em><em>O</em><em>)</em><em>=</em><em>-</em><em>2</em>

<em>2</em><em> </em><em>(</em><em>S</em><em>)</em><em> </em><em>+</em><em> </em><em>3</em><em>(</em><em>-</em><em>2</em><em>)</em><em>=</em><em>-</em><em>2</em>

<em>2</em><em> </em><em>(</em><em>S</em><em>)</em><em> </em><em>+</em><em> </em><em>(</em><em>-</em><em>6</em><em>)</em><em> </em><em>=</em><em>-</em><em>2</em>

<em>2</em><em> </em><em>S</em><em> </em><em>=</em><em> </em><em>+</em><em>4</em>

<em>S</em><em>=</em><em> </em><em>+</em><em /><em>2</em><em />

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Explanation:

From the given information:

The equation for the reaction can be represented as:

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The I.C.E table can be represented as:

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Initial:             14                  2.6                     0

Change:        -2x                -x                      +2x

Equilibrium:   14 - 2x          2.6 - x                2x

However, Since the amount of sulfur trioxide gas to be 1.6 mol.

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For 2SO₂; we have 14 - 2x

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= 14 - 1.6

= 12.4 mol

For O₂; we have 2.6 - x

= 2.6 - 1.6

= 1.0 mol

Thus;

[SO₂] = moles / volume = ( 12.4/50) = 0.248 M ,

[O₂] = 1/50 = 0.02 M ,  

[SO₃] = 1.6/50 = 0.032 M

Kc = [SO₃]² / [SO₂]² [O₂]

= ( 0.032²) / ( 0.248² x 0.02)

= 0.8325

Recall that; the equilibrium constant for the reaction 2SO_2 + O_2 \to 2SO_3 = 0.8325;

If we want to find:

SO_2 + \dfrac{1}{2}O_2 \to SO_3

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K_c = (0.8325)^{1/2}

\mathbf{K_c = 0.912}

Since no temperature is given to use in the question, it will be impossible to find the final temperature of the mixture.

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▹ Answer

<em>x¹⁰z¹⁵</em>

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━━━━━━━☆☆━━━━━━━

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