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Andrej [43]
2 years ago
14

A + B = AB

Chemistry
1 answer:
Ahat [919]2 years ago
7 0

The greatest amount of AB would be produced if the equilibrium constant of the reaction is equal to 1.0 \;X \;10^5. Hence, option D is correct.

<h3>What is an equilibrium constant?</h3>

A number that expresses the relationship between the amounts of products and reactants present at equilibrium in a reversible chemical reaction at a given temperature.

The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants.

If the value of K is greater than 1, the products in the reaction are favoured. If the value of K is less than 1, the reactants in the reaction are favoured.

Hence, option D is correct.

Learn more about the equilibrium constant here:

brainly.com/question/10038290

#SPJ1

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04.05 mol
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Answer:

34.8 g

Explanation:

Answer:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:         123.90    70.91      208.24

                 P₄   +   20Cl₂ ⟶ 4PCl₅

Mass/g:   46.0       32.0

2. Calculate the moles of each reactant  

\text{moles of P}_{4} = \text{46.0 g P}_{4} \times \dfrac{\text{1 mol P}_{4}}{\text{123.90 g P}_{4}} = \text{0.3713 mol P}_{4}\\\\\text{moles of Cl}_{2} = \text{32.0 g Cl }_{2} \times \dfrac{\text{1 mol Cl }_{2}}{\text{70.91 g Cl }_{2}} = \text{0.4513 mol Cl }_{2}

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The molar ratio is 4 mol PCl₅:4 mol P₄

\text{Moles of PCl}_{5} = \text{0.3713 mol P}_{4} \times \dfrac{\text{4 mol PCl}_{5}}{\text{4 mol P}_{4}} = \text{0.3713 mol PCl}_{5}

From Cl₂:

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\text{Moles of PCl}_{5} =  \text{0.4513 mol Cl}_{2}\times \dfrac{\text{4 mol PCl}_{5}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol PCl}_{5}

4. Identify the limiting and excess reactants

The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.

The excess reactant is phosphorus.

5. Mass of excess reactant

(a) Moles of P₄ reacted

The molar ratio is 1 mol P₄:20 mol Cl₂

\text{Moles reacted} = \text{0.4513 mol Cl}_{2} \times \dfrac{\text{4 mol P}_{4}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol P}_{4}

(b) Mass of P₄ reacted

\text{Mass reacted} = \text{0.090 26 mol P}_{4} \times \dfrac{\text{123.90 g P}_{4}}{\text{1 mol P}_{4}} = \text{11.18 g P}_{4}

(c) Mass of P₄ remaining

Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄

4 0
3 years ago
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meriva

Answer:

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Explanation:

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Answer:

The correct answer is

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