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4 years ago

CH, (g) + 20 (g) → CO(g) + 2H 0(g)

Chemistry

1 answer:

galben [10]4 years ago

6 0

<h3>Answer:</h3>

126.14 g

<h3>Explanation:</h3>

Complete combustion of hydrocarbons yields water and carbon dioxide.
Methane is a hydrocarbon in the homologous series known as alkanes.
Methane undergoes combustion in air to produce water and carbon dioxide according to the equation below.

CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(g)

<u>We are given;</u>

3.5 moles of Methane

We are required to determine the mass of H₂O produced

<h3>Step 1: Moles of H₂O produced</h3>

From the equation 1 mole of methane undergoes combustion to produce 2 mole of H₂O.

Therefore, Moles of methane = Moles of H₂O × 2

Hence, Moles of H₂O = 7.0 moles

<h3>Step 2: Mass of H₂O produced </h3>

We know that; mass = Moles × Molar mass

Molar mass of water = 18.02 g/mol

Therefore;

Mass of water = 7.0 moles × 18.02 g/mol

= 126.14 g

Thus, the mass of water produced is 126.14 g

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This is an incomplete question, here is a complete question and an image is attached below.

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In mixture 1 : There are 5 $NO$ and 4 $O_2$ molecules.

In mixture 2 : There are 7 $NO$ and 2 $O_2$ molecules.

In mixture 3 : There are 3 $NO$ and 5 $O_2$ molecules.

Now we have to determine the rate law expression for mixture 1, 2 and 3.

The rate law expression for mixture 1 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(5)^2\times (4)$

$Rate=k(100)$

The rate law expression for mixture 2 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(7)^2\times (2)$

$Rate=k(98)$

The rate law expression for mixture 3 is:

$Rate=k[NO]^2[O_2]$

$Rate=k(3)^2\times (5)$

$Rate=k(45)$

Hence, the mixture 1 has the fastest initial rate.

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Answer:

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