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sergiy2304 [10]
3 years ago
12

Write a balanced equation for the complete oxidation reaction that occurs when ethane burns in air

Chemistry
1 answer:
damaskus [11]3 years ago
6 0
<h3>Balanced equation : 2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)</h3><h3>Further explanation</h3>

Alkanes are saturated hydrocarbons that have single bonds in chains

General formula for alkanes :

\tt \large{\bold{C_nH_{2n+2}}

Hydrocarbon combustion reactions (specifically alkanes)

\large {\box {\bold{C_nH _ (_2_n _ + _ 2_) + \dfrac {3n + 1} {2} O_2 \Rightarrow nCO_2 + (n + 1) H_2O}}}

So that the burning of ethane with air (oxygen):

\tt C_2H_6+\dfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O

2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (ℓ)

or we can use mathematical equations to solve equilibrium chemical equations by giving the coefficients for each compound involved in the reaction

C₂H₆ (g) + aO₂ (g) ⟶ bCO₂ (g) + cH₂O (ℓ)

C : left 2, right b ⇒ b=2

H: left 6, right 2c⇒ 2c=6⇒ c= 3

O : left 2a, right 2b+c⇒ 2a=2b+c⇒2a=2.2+3⇒2a=7⇒a=7/2

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When heated, KClO3 decomposes into KCl and O2. KClO3--&gt; 2KCl+ 3O . If this reaction produced 62.6 g of KCl, how much O2 was p
Vladimir [108]
The solution to your problem is as follows:

 <span>2 KClO3 → 2 KCl + 3 O2 

</span><span>MW of KCL = 39.1 + 35.35 = 74.6 g/mole 

62.6/74.6 = 0.839 moles KCl produced 

0.839 moles KCl x 3O2/2KCl x 32g/mole O2 = 40.27g of O2 was produced
</span>
Therefore, there are <span>40.27g of O2 produced during the reaction.
</span>
I hope my answer has come to your help. Thank you for posting your question here in Brainly.


5 0
3 years ago
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brilliants [131]

The right answer is

Table A Organic solvent

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6 0
3 years ago
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If the radius of copper is 0.128 nm, what is the pd of the (100) plane for copper in m-2?
kolbaska11 [484]

Copper has a FCC i.e. face centered cubic crystal structure. The 100 plane is essentially a planar section of the cubic cell where 4 Cu atoms occupy the 4 corners of the plane along with 1 Cu atom at the center of that plane. Each of the Cu atoms in the corners is shared by 4 adjacent unit cells. Thus, there are 2 Cu atoms present in the 100 plane (4*1/4 + 1 = 2).

Now, the planar density PD along the 100 plane is given as:

PD(100) = # atoms in the 100 plane/Area of 100 plane

            = \frac{2}{8R^{2} }

Here R = radius = 0.128 nm = 0.128*10^{-9} m

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6 0
3 years ago
Match Term Definition
mart [117]

The correct answers are:

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7 0
3 years ago
How do the percent compositions for C3H6 and C4H7 compare?
mariarad [96]

A. They are the same

<h3>Further explanation</h3>

Given

C3H6 and C4H8

Required

The percent compositions

Solution

  • C₃H₆(MW = 42 g/mol)

%C = 3.12/42 x 100% = 85.71%

%H = 6.1/42 x 1005 = 14.29%

C₄H₈(MW=56 g/mol)

%C = 4.12/56 x 100% = 85.71%

%H = 8.1/56 x 100%=14.29%

So they are the same, because mol ratio of C and H in both compounds is the same, 1: 2

3 0
3 years ago
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