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shepuryov [24]
2 years ago
15

In the oxidation of ethane: 2 C2H6 + 7 02 → 4 CO2 + 6 H2O how many

Chemistry
1 answer:
Vera_Pavlovna [14]2 years ago
5 0

Answer:

10.5 L of O₂.

Explanation:

We'll begin by calculating the number of mole in 90 g of C₂H₆. This can be obtained as follow:

Molar mass of C₂H₆ = (12×2) + (6×1)

= 24 + 6

= 30 g/mol

Mass of C₂H₆ = 90 g

Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 90 / 30

Mole of C₂H₆ = 3 moles.

Finally, we shall determine the volume of O₂ required. This is illustrated:

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above, we can say that:

2 L of C₂H₆ required 7 L of O₂.

Therefore, 3 L of C₂H₆ will require

= (3 × 7)/ 2 = 10.5 L of O₂.

Thus, 10.5 L of O₂ is required for the reaction.

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Reversible reaction and examples on them
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Answer:

Here are three examples  

Explanation:

In a reversible reaction, the conversions of reactants to products and of products to reactants occur at the same time.

Example 1

The reaction of hydrogen and iodine to from hydrogen iodide.  

H₂ + I₂ ⇌ 2HI

Example 2

The dissociation of carbonic acid in water to form hydronium and hydrogen carbonate ions

H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻

Example 3

The dissociation of dinitrogen tetroxide to nitrogen dioxide.

N₂O₄ ⇌ 2NO₂

4 0
3 years ago
A balloon is filled with 0.75 L of Helium gas at 35 °C. If the temperature is increased to 113 °C, what thewill new volume be?
faust18 [17]

Answer: 0.9398 Liters

Explanation:

Charles' Law

V1/ T1 = V2/T2

0.75/35 = ?/113

0.75/35 = 0.9398/113

4 0
3 years ago
The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t
Andrews [41]

<u>Answer:</u> The experimental van't Hoff factor is 1.21

<u>Explanation:</u>

The expression for the depression in freezing point is given as:

\Delta T_f=iK_f\times m

where,

i = van't Hoff factor = ?

\Delta T_f = depression in freezing point  = 0.225°C

K_f = Cryoscopic constant  = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21

Hence, the experimental van't Hoff factor is 1.21

7 0
3 years ago
How much kilometers is in 6000m
lord [1]

Answer:

about 3 miles i think

Explanation:

3 0
2 years ago
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Why is it better to give the actual speed of an object rather than to simply say it moved fast or it moved slowly
DIA [1.3K]
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3 years ago
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