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3 years ago

How many moles of hydrogen gas would be needed to react with excess carbon dioxide to produce 88.1 molesmoles of water vapor?

Chemistry

1 answer:

irga5000 [103]3 years ago

8 0

Answer:

88.1 moles of H₂

Explanation:

Let's make the reaction:

H₂(g) + CO₂(g) → H₂O(g) + CO(g)

Ratio is 1:1

Therefore, 1 mol of hydrogen reacts with 1 mol of carbon dioxide to produce 1 mol of water and 1 mol of carbon monoxide.

In conclusion, 88.1 moles of water must be produced by 88.1 moles of H₂

The reaction can also be written as an equilbrium

H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g) Kc

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murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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The correct answer would be A

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In a room full of air, the air is mainly composed of Nitrogen and Oxygen molecules (both at room temperature). Find (to two sign

zloy xaker [14]

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$v_{rms} = \sqrt{\frac{3kT}{M}}$

where, $v_{rms}$ = root mean square speed

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svet-max [94.6K]

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