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JulijaS [17]
3 years ago
6

Consider the balanced chemical equation that follows. You are asked to determine how many moles of water you can produce from 4.

0 mol of hydrogen and excess oxygen. (Excess oxygen means that so much oxygen is available it will not run out.) Which of the numbers that appear in the balanced chemical equation below are used to perform this calculation?
Chemistry
1 answer:
matrenka [14]3 years ago
3 0

Explanation:

According to law of conservation of mass, mass of the reactants is equal to the mass of products in a chemical equation. As mass can neither be created nor it can be destroyed but it can be transformed from one form to another.

As it is given that hydrogen and in excess oxygen is reacting that leads to the formation of water. Hence, the chemical reaction equation will be as follows.

           2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)

Since, it is given that 4 mol of hydrogen is reacting with excess of oxygen and gives 2 moles of water.

Hence, number of moles of water produced is calculated as follows.

              4 mol of H_{2} \times \frac{2 mol of H_{2}O}{2 mol of H_{2}}

                  = 4 moles of H_{2}O

Thus, we can conclude that 4 moles of water you can produce from 4.0 mol of hydrogen and excess oxygen.

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oksian1 [2.3K]
I think the answer is 1.31m

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4 0
3 years ago
The water-gas shift reaction describes the reaction of carbon monoxide and water vapor to form carbon dioxide and hydrogen (the
sesenic [268]

Answer:

ΔH∘ = - 41.2 KJ

Explanation:

We want to obtain the change in enthalpy for the reaction

CO(g) + H₂O(g) → CO₂(g) + H₂(g) (Main reaction)

And we're given the heat of formation of the reactants and products in the reaction

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

To achieve this, we use the Born-Haber cycle.

The Born-Haber cycle entails writing the change in enthalpy of a reaction as a sum of change in enthalpies of a number of reactions that sum up to give the reaction whose enthalpy we needed from the start.

The main reaction is a sum of a sort of combination of Reactions A, B and C. We find this combination now.

From the reactions whose change in enthalpies are given,

C(s) + O₂(g) → CO₂(g) ΔH∘=−393.5kJ (Reaction A)

2CO(g) → 2C(s) + O₂(g) ΔH∘=+221.0kJ (Reaction B)

Dividing through by 2

CO(g) → C(s) + (1/2)O₂(g) ΔH∘=+110.5kJ (the enthalpy is divided by 2 too)

This reaction becomes (Reaction B)/2

2H₂(g) + O₂(g) → 2H₂O(g) ΔH∘=−483.6kJ (Reactions C)

Changing the direction of the reaction

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=483.6kJ (the sign on the change in enthalpy changes)

Then, dividing by 2

2H₂O(g) → 2H₂(g) + O₂(g) ΔH∘=+483.6kJ

H₂O(g) → H₂(g) + (1/2)O₂(g) ΔH∘=241.8kJ (the change in enthalpy is divided by 2 too)

This reaction becomes (-Reaction C)/2

But, now, our main reaction can be written as a sum of these new Reactions,

Main Reaction = (Reaction A) + [(Reaction B)/2] + [(- Reaction C)/2]

C(s) + O₂(g) + CO(g) + H₂O(g) → CO₂(g) + C(s) + (1/2)O₂(g) + H₂(g) + (1/2)O₂(g)

Which gives the main reaction after eliminating the O2 that appears on both sides.

CO(g) + H₂O(g) → CO₂(g) + H₂(g)

Hence,

(ΔH∘ for the main reaction) = (ΔH∘ for reaction A) + [(ΔH∘/2) for reaction B) - [(ΔH∘/2) for reaction C)

ΔH∘ = - 393.5 + (221/2) - (-483.6/2) = - 41.2 KJ

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3 years ago
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lions [1.4K]
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142 g ---------------- 6.02 x 10²³ molecules
0.920g g ------------ ( molecules )

molecules = 0.920 x ( 6.02 x 10²³ ) / 142

molecules = 5.53 x 10²³ / 142

= 3.89 x 10²¹ molecules

1 molecule P2O5 -------------------------- 7 atoms
3.89 x 10²¹ molecules -------------------- ( atoms )

atoms = ( 3.89 x 10²¹) x 7 / 1

atoms = 2.72 x 10²² atoms of P2O5

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Answer:

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