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Mars2501 [29]
3 years ago
10

Assign oxidation numbers for the following:

Chemistry
1 answer:
Readme [11.4K]3 years ago
6 0

Answer:

1. P₂O₅      → oxidation number of phosphorous is +5 and Oxygen is -2.

2. (SO₄)²⁻ → oxidation number of sulfur is +6 and Oxygen is -2.

3. KClO₃   → oxidation number of Potassium is +1, Chlorine is +5, and Oxygen is -2.

4. NH₄Cl  → oxidation number of Nitrogen is -3, Hydrogen is +1, and Chlorine is -1

5. (NH₄)₂S  → oxidation number of Nitrogen is -3, Hydrogen is +1, and Sulfur is -2

Explanation:

General Rules for assigning oxidation numbers

The oxidation number of a free element is always 0.

The oxidation number of a mono-atomic ion equals the charge of the ion.

The alkali metals (group I) always have an oxidation number of +1.

The alkaline earth metals (group II) are always assigned an oxidation number of +2.

Oxygen almost always has an oxidation number of -2, except in peroxides (H₂O₂) where it is -1 and in compounds with fluorine (OF₂) where it is +2.

Hydrogen has an oxidation number of +1 when combined with non-metals, but it has an oxidation number of -1 when combined with metals.

The algebraic sum of the oxidation numbers of elements in a compound is zero.

The algebraic sum of the oxidation states in an ion is equal to the charge on the ion.

Using the above rules:

<u>1. P₂O₅</u>

∵ it is a neutral compound its total charge is 0.

Also, we know that oxygen has an oxidation number of -2.

Let oxidation number of P be x

∴ 2(x)+5(-2)=0   →   2x=+10   →   x=+5  

∴oxidation number of phosphorous is +5.

<u>2. SO₄²⁻:</u>

∵ it is a charged ion its total charge is -2.

Also, we know that oxygen has an oxidation number of -2.

Let oxidation number of S be x

∴ (x)+4(-2)= -2   →   x=+6  

∴oxidation number of sulfur is +6.

<u>3. KClO₃:</u>

∵ it is a neutral compound its total charge is 0.

Also, we know that oxygen has an oxidation number of -2 and the  oxidation number of K (group I) is +1

Let oxidation number of Cl be x

∴ (+1) + (x) + 3(-2) = 0   →   x=+5  

∴oxidation number of Chlorine is +5.

<u>4. NH₄Cl:</u>

∵ it is a neutral compound its total charge is 0.

Also, we know that chloride has an oxidation number of -1

Hydrogen has an oxidation number of +1 when combined with non-metals

 Let oxidation number of N be x

∴ (x) + 4(+1) + (-1) = 0   →   x=-3  

∴oxidation number of Nitrogen is -3.

<u>5. (NH₄)₂S:</u>

∵ it is a neutral compound its total charge is 0.

Also, we know that chloride has an oxidation number of -1

Ammonium ion (NH₄⁺) has an oxidation number of +1

 Let oxidation number of N be x

∴ 2(+1) + (x) = 0   →   x= -2  

∴oxidation number of sulfur is -2.

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Answer:

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Explanation:

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En este caso, dado que tenemos la siguiente reacción química, la cual se puede balancear directamente:

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Podemos calcular la masa máxima de cualquier producto, digamos PH3, al comparar la masa de este, que 2 g the Ca3P2 y 1 g de H2O producen por separado y de acuerdo a la estequiometría:

2gCa_3P_2 *\frac{1molCa_3P_2 }{182.18gCa_3P_2 }*\frac{2molPH_3}{1molCa_3P_2 }  *\frac{34gPH_3}{1molPH_3 } =0.747gPH_3\\\\1gH_2O *\frac{1molH_2O }{18.02gH_2O}*\frac{2molPH_3}{6molH_2O}  *\frac{34gPH_3}{1molPH_3 } =0.629gPH_3

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