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3 years ago

How to balance CH4 + O2=CO2+H2O

Chemistry

2 answers:

sertanlavr [38]3 years ago

7 0

Answer:

LHS-RHS

1 C - 1 C

4 H-2 H

2 O- 3 O

So on right side 2 hydrogen are less and one oxygen is more..so

Balanced equation is

CH4+2O2==CO2 + 2H2O

Explanation:

Anastasy [175]3 years ago

6 0

Hey there!

CH₄ + O₂ → CO₂ + H₂O

Balance H.

4 on the left, 2 on the right. Add a coefficient of 2 in front of H₂O.

CH₄ + O₂ → CO₂ + 2H₂O

Balance C.

1 on the left, 1 on the right. Already balanced.

Balance O.

2 on the left, 4 on the right. Add a coefficient of 2 in front of O₂.

CH₄ + 2O₂ → CO₂ + 2H₂O

Our final balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O

Hope this helps!

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The equilibrium constant for the reaction 2NO(g)+Br2(g)⥫⥬==2NOBr(g) is Kc=1.3×10−2 at 1000 K. At this temperature does the equil

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The equilibrium favors NO and $Br_2$ .

(1) The value of equilibrium constant for this reaction is, 76.9

(2) The value of equilibrium constant for this reaction is, 8.77

Explanation:

The given chemical equation is:

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

The value of equilibrium constant for the above equation is $K_c=1.3\times 10^{-2}$ .

The value of $K_c$ that means equilibrium lies to the left side. Thus, the equilibrium favors NO and $Br_2$ .

We need to calculate the equilibrium constant for the given equation of above chemical equation, which is:

(1) $2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_1}=\frac{1}{K_c}$

$K_{c_1}=\frac{1}{1.3\times 10^{-2}}=76.9$

Thus, the value of equilibrium constant for this reaction is, 76.9

(2) $NOBr(g)\rightarrow NO(g)+\frac{1}{2}Br_2(g)$

The equilibrium constant for the reverse reaction will be the reciprocal of the initial reaction.

If the equation is multiplied by a factor of '1/2', the equilibrium constant of the reverse reaction will be the 1/2 power of the equilibrium constant of initial reaction.

The value of equilibrium constant for reverse reaction is:

$K_{c_2}=(\frac{1}{K_c})^{1/2}$

$K_{c_2}=(\frac{1}{1.3\times 10^{-2}})^{1/2}=8.77$

Thus, the value of equilibrium constant for this reaction is, 8.77

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