What is the chemical formula for 8.6 mol of sulfur and 3.42 mol of phosphorus

Chemistry

1 answer:

adell [148]1 year ago

5 0

The chemical formula for the compound containing 8.6 mol of sulfur and 3.42 mol of phosphorus is P₂S₅

<h3>How do I determine the formula of the compound?</h3>

From the question given above, the following data were obatined:

Sulphur (S) = 8.6 moles
Phosphorus (P) = 3.42 mole
Chemical formula =?

The chemical formula of the compound can be obtained as follow:

Divide by their molar mass

S = 8.6 / 32 = 0.26875

P = 3.42 / 31 = 0.11032

Divide by the smallest

S = 0.26875 / 0.11032 = 2.44

P = 0.11032 / 0.11032 = 1

Multiply by 2 to express in whole number

S = 2.44 × 2 = 5

P = 1 × 2 = 2

Thus, the chemical formula is P₂S₅

Learn more about empirical formula:

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Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)

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Explanation:

The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.

Suppose we have the following half-reactions.

Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V

Zn²⁺(⁺aq) + 2 e⁻ → Zn(s) E°red = -0.76 V

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻ E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

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