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3 years ago

Consider a situation in which 211 g

Chemistry

1 answer:

Stella [2.4K]3 years ago

3 0

Answer:

3.00 mol

Explanation:

Given data:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂ → 2P₂O₅

Number of moles of P₄:

Number of moles = mass/ molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂ :

Number of moles = mass/ molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

O₂ : P₂O₅

5 : 2

7.5 : 2/5×7.5 = 3.00

P₄ : P₂O₅

1 : 2

1.7 : 2×1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles × molar mass

Mass = 3 mol ×283.9 g/mol

Mass = 852 g

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34kurt

Answer:

For a: The balanced chemical equation is written below.

For b: The corrosion of iron pipe will take place in the presence of zinc.

For c: Zinc will not protect iron pipe from corrosion.

Explanation:

For a:

The given half reaction follows:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, zinc will undergo reduction reaction will get reduced.

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.763V$

Reduction half reaction: $Fe+2e^-\rightarrow Fe;E^o_{Fe^{2+}/Fe}=-0.441V$

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For b:

For a reaction to be spontaneous (thermodynamically feasible) , the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

Calculating the $E^o_{cell}$ using above equation, we get:

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cestrela7 [59]

Answer:

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