Answer:
2C₂H₆ + [7]O₂ → [4]CO₂ + [6]H₂O
Explanation:
Chemical equation:
C₂H₆ + O₂ → CO₂ + H₂O
Balanced chemical equation:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Step 1:
2C₂H₆ + O₂ → CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 1
H = 12 H = 2
O = 2 O = 3
Step 2:
2C₂H₆ + O₂ → 4CO₂ + H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 2
O = 2 O = 9
Step 3:
2C₂H₆ + O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 2 O = 14
Step 4:
2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O
Left hand side Right hand side
C = 4 C = 4
H = 12 H = 12
O = 14 O = 14
Kilometers, Meters and centimeters if metric
Feet, inches, yards and miles if customary ( u.s.)
Complete Question
The complete question is shown on the first uploaded image
Answer:
The equilibrium constant is 
Explanation:
From the question we are told that
The chemical reaction equation is

The voume of the misture is
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
The molar mass of
is a constant with value of 
Generally the number of moles is mathematically given as

For 


For 


For 


Generally the concentration of a compound is mathematicallyrepresented as

For 
![Concentration[Fe_2 O_3] = \frac{0.222125}{5.4}](https://tex.z-dn.net/?f=Concentration%5BFe_2%20O_3%5D%20%3D%20%5Cfrac%7B0.222125%7D%7B5.4%7D)
For 
![Concentration[H_2] = \frac{1.815}{5.4}](https://tex.z-dn.net/?f=Concentration%5BH_2%5D%20%3D%20%5Cfrac%7B1.815%7D%7B5.4%7D)

For 
![Concentration [H_2O] = \frac{0.12}{5.4}](https://tex.z-dn.net/?f=Concentration%20%5BH_2O%5D%20%3D%20%5Cfrac%7B0.12%7D%7B5.4%7D)

The equilibrium constant is mathematically represented as
![K_c = \frac{[concentration \ of \ product]}{[concentration \ of \ reactant ]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cfrac%7B%5Bconcentration%20%5C%20of%20%5C%20product%5D%7D%7B%5Bconcentration%20%5C%20of%20%5C%20reactant%20%5D%7D)
Considering 
And 
At equilibrium the


Answer:
At equilibrium, the concentration of
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no
nor
present. Immediately,
and
are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: [
]=0 ; [
]= 0 ; [
]=0.60M
C: [
]=+x ; [
]= +x ; [
]=-2x
E: [
]=0+x ; [
]= 0+x ; [
]=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 




At equilibrium, the concentration of
is going to be 0.30M