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VARVARA [1.3K]
3 years ago
14

What volume (in L) of oxygen will be required to produce 77.4 L of water vapor in the reaction below?

Chemistry
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

For the production of 77.4 L water 90.3 L oxygen is required.

Explanation:

Given data:

Volume of oxygen required = ?

Volume of water produced = 77.4 L

Solution:

Chemical reaction equation:

2C₂H₆ + 7O₂  →  4CO₂ + 6H₂O

1 mole = 22.414 L

There are 6 moles of water = 6×22.414 = 134.5 L

There are 7 moles of oxygen = 7×22.414 = 156.9 L

Now we will compare the litters of water and oxygen:

                              H₂O           :              O₂    

                              134.5         :              156.9

                                77.4         :             156.9/134.5×77.4 =90.3 L

So for the production of 77.4 L water 90.3 L oxygen is required.

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Explanation:

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This is how fluorine appears in the periodic table.9 F Flourine 19.00 What information does “9” give about an atom of fluorine?
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The atomic number, the number of protons and the number of electrons.
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How many sulfur atoms are present in 25.6 g of Al2(S2O3)3
IgorC [24]

Given the mass of Al_{2}(S_{2}O_{3})_{3}=25.6 g

The molar mass of Al_{2}(S_{2}O_{3})_{3}=390.35g/mol

Converting mass of Al_{2}(S_{2}O_{3})_{3}to moles:

25.6 g Al_{2}(S_{2}O_{3})_{3}*\frac{1molAl_{2}(S_{2}O_{3}}{390.35 gAl_{2}(S_{2}O_{3}} =0.0656molAl_{2}(S_{2}O_{3}

Converting mol Al_{2}(S_{2}O_{3})_{3}to mol S:

0.0656mol Al_{2}(S_{2}O_{3})_{3}*\frac{6molS}{1mol Al_{2}(S_{2}O_{3})_{3}}=0.3936 molS

Converting mol S to atoms of S using Avogadro's number:

1 mol = 6.022*10^{23}atoms

0.3936mol S *\frac{6.022*10^{23}atoms S}{1 mol S}=2.37*10^{23} S atoms

5 0
3 years ago
Keq for the reaction below is 2400. If the initial conditions of the reaction are a 1.0 L flask that contains 0.024 mol NO (g),
podryga [215]

Answer:

The answer to your question is it is not at equilibrium, it will move to the products.

Explanation:

Data

Keq = 2400

Volume = 1 L

moles of NO = 0.024

moles of N₂ = 2

moles of O₂ = 2.6

Process

1.- Determine the concentration of reactants and products

[NO] = 0.024 / 1 = 0.024

[N₂] = 2/1 = 2

[O₂] = 2.6/ 1= 2.6

2.- Balanced chemical reaction

                     N₂ + O₂    ⇒   2NO

3.- Write the equation for the equilibrium of this reaction

                     Keq = [NO]²/[N₂][O₂]

- Substitution

                    Keq = [0.024]² / [2][2.6]

-Simplification

                    Keq = 0.000576 / 5.2

-Result

                    Keq = 1.11 x 10⁻⁴

Conclusion

It is not at equilibrium, it will move to the products because the experimental Keq was lower than the Keq theoretical-

                         1.11 x 10⁻⁴ < 2400

7 0
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Inessa [10]

Answer:

I think the answer is C- radiation

Explanation:

If its not C its A Hope this helps forgive me if i'm wrong :/

8 0
3 years ago
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