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AlekseyPX
3 years ago
14

4HCl(g)+O2(g)⟶2Cl2(g)+2H2O(g) Calculate the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2. mass:

g
Chemistry
1 answer:
FromTheMoon [43]3 years ago
6 0

The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 10^{23} of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Moles of hydrochloric acid = 0.385 mol

Mass of chlorine gas =?

Chemical equation:

4HCl +  O₂ → 2Cl₂  + 2H₂O

Now we will compare the moles of Cl₂ with HCl.

                 HCl           :               Cl₂

                   4             :               2

                0.385       :              2÷4× 0.385 = 0.1925 mol

Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.

Mass of Cl₂ :

Mass of Cl₂ = moles × molar mass

Mass of Cl₂ =0.1925 mol × 71 g/mol

Mass of Cl₂ =  13.6675 g

Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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Answer:

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Explanation:

Given data:

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c = 125520 J /  0.0228 g.°C

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<h3>Concentration of each gas</h3>

CO2 = 2 mol/5 L = 0.4

H2 = 1.5 mol/5L = 0.3

<h3>ICE table</h3>

Create ICE table as shown below

        CO2(g)    +     H2(g) ↔       CO(g)    +     H2O(g)

I          0.4               0.3                0                  0

C        - x                 - x                 x                   x

E        0.4 - x           0.3 - x           x                   x

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x² = 2.5(0.4 - x)(0.3 - x)

x² = 2.5(0.12 - 0.7x + x²)

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solve the quadratic equation using formula method;

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Thus, the concentration of carbon monoxide at equilibrium is 0.209 M.

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