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3 years ago

How many grams of o2 are required to produce 358.5 grams of zno? 2zn + o2 ® 2zno?

1 answer:

6 0

The balanced equation for the reaction is ;
2Zn + O2 —> 2ZnO
The stoichiometry of O2 to ZnO is 1:2
The mass of ZnO formed - 358.5 g
The number of moles formed - 358.5 g / 81.4 g/mol = 4.4 moles
Therefore number of O2 moles reacted = 4.4 moles /2 = 2.2 mol
Mass of O2 reacted = 2.2 mol x 32 g/mol = 70.4 g

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Hello, here’s the answer to your question. Converting ammonia to nitrate, which is absorbed by plants

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2 years ago

Describe how particles of iron (Fe) an oxygen (O2) react to how produce oxide (Fe2O3), also known as rust.

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Answer:

Described by a redox reaction below

Explanation:

Iron(III) oxide is an ionic compound, since it consists of a metal, iron, and a nonmetal, oxygen.

Ionic compounds are formed when metals lose their valence electrons in order to have an octet in their previous shell and donate them to nonmetal atoms, so that nonmetals fill their outer shell to have an octet.

As a result, positive ions (cations) and negative ions (anions) are formed. When iron reacts with oxygen, the following reaction takes place:

$4 Fe (s) + 3 O_2 (g)\rightarrow 2 Fe_2O_3 (s)$

This is a redox (oxidation–reduction) reaction, since we have electron loss and gain. Four iron atoms lose a total of 12 electrons to obtain a +3 charge in the final compound, while 3 oxygen molecules gain these 12 electrons to become 6 oxide anions with a -2 charge.

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3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

ruslelena [56]

Answer:

b. 16, reactant side

Explanation:

Let's consider the following redox reaction.

MnO₄⁻(aq) + I⁻(aq) → Mn²⁺(aq) + I₂(s)

We can balance it using the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: I⁻(aq) → I₂(s)

Step 2: Perform the mass balance, adding H⁺(aq) and H₂O(l) where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s)

Step 3: Perform the charge balance, adding electrons where appropriate

MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 I⁻(aq) → I₂(s) + 2 e⁻

Step 4: Multiply both half-reactions by numbers so that the number of electrons gained and lost are equal

2 × (MnO₄⁻(aq) + 8 H⁺(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 I⁻(aq) → I₂(s) + 2 e⁻)

Step 5: Add both half-reactions and cancel what is repeated on both sides

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 e⁻ + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s) + 10 e⁻

The balanced reaction is:

2 MnO₄⁻(aq) + 16 H⁺(aq) + 10 I⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 I₂(s)

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2 years ago

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Answer:

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3 years ago

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Answer:

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