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Diano4ka-milaya [45]
2 years ago
9

A mixture of 2.0 mol of CO(g)? and 2 mol of H2O?(g)? was allowed to come to equilibrium in a 1.0L flask at a high temperature. I

f Kc?=4.0, what is the molar concentration of H2(g) in the equilibrium mixture: CO(g) + H2O(g)\rightleftharpoons CO2(g) +H2(g)
Chemistry
1 answer:
Bess [88]2 years ago
7 0

For a mixture of 2.0 mol of CO(g) and 2 mol of H2O(g) was allowed to come to equilibrium in a 1.0L flask at a high temperature. the molar concentration of H2(g)  is mathematically given as

[H2] = x = 1.33 M

<h3>What is the molar concentration of H2(g) in the equilibrium mixture?</h3>

Generally, the equation for the Chemical reaction  is mathematically given as

Kc = [CO2]*[H2]/[CO]*[H2O]

Therefore

4.0 = (1*x)^2/(2-1*x)^2

x = 1.33

In conclusion, the molar concentration

[H2] = x = 1.33 M

Read more about  Chemical reaction

brainly.com/question/16416932

#SPJ1

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Explanation:

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8 0
3 years ago
Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin
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Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

                       CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

                    CH₄     +     4Cl₂ --------→ CCl₄ + 4HCl

                    1 mol        4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of  Cl₂  = 71 g/mol

mass of Cl₂

                mass in grams = no. of moles x molar mass

                mass of Cl₂ = 4 mol x 71 g/mol

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mass of CH₄

                mass in grams = no. of moles x molar mass

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So,

284 g of Cl₂  combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂  

Apply unity Formula

                           284 g of Cl₂  ≅ 16 g of methane ( CH₄ )

                           33.4 g of Cl₂  ≅ X g of methane ( CH₄ )

By cross multiplication

                          X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

                          X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

So

methane (CH₄) = 1.882 g

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