Which of the following is true about the following chemical reaction at equilibrium? PICK ALL THAT APPLY.

Chemistry

1 answer:

Alekssandra [29.7K]2 years ago

4 0

Answer:

Option B and C are correct

Explanation:

Chemical reaction at equillibrium means: the rate of forward reaction equals the rate of reverse reaction. This means no product will be consumed, neither formed.

A) The concetration of H2O equals the concentration of O2. This is false.

At the equillibrium the concetnrations are constant. But not necessarily equal.

B) The rate of the forward reaction, reactants to products, equals the rate of the reverse reaction, products to reactants. This is true. IT's the definition of a the equillibrium.

C) The C-C bonds hold more energy in C6H12O6 than the C-O bonds in CO2. Hint: think teter-toter.

This is to, C-C bonds hold more energy than C-O, that's why C6H12O6 is burned, energy is released.

D) The concentration of O2 is changing at equilibrium. This is false. The concentration of O2 is constant.

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Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts

SOVA2 [1]

Answer:

$\boxed{\text{Mg is the limiting reactant}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble the data in one place.

2Mg + O₂ ⟶ 2MgO

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Calculate the moles of MgO we can obtain from each reactant.

From Mg:

The molar ratio of MgO:Mg is 2:2

$\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}$

From O₂:

The molar ratio of MgO:O₂ is 2:1.

$\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}$