Question

Identify the type of reaction and balance the following equation:
CH4 + O2 →CO2 + H2O + heat

Answer 1

Answer:

Balanced equation-CH4 + 2O2 → CO2 + 2H2O

reaction name-H2 Nano Hydrogen Water

Bottle

Explanation:

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Answer 2

Answer:

• Given reaction is Exothermic Reaction.

• $\displaystyle\sf CH_{4} + 2\;O_{2 }\longrightarrow CO_{2} + 2\;H_{2}O+Heat$

Explanation:

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As from the skeletal equation we can see that the heat is liberated on the product side, therefore the reaction is Exothermic.

Now, Balancing the given equation,

The given Reaction is ,

$\displaystyle\sf CH_{4} + O_{2 }\longrightarrow CO_{2} + H_{2}O+Heat$

Now, Checking the initial & final atoms of Each element.

$\large\begin{tabular}{|c|c|c|c|}\cline{1-4}Element&C&O&H\\\cline{1-4} Initial&1&2&4\\\cline{1-4}Final&1&3&2\\\cline{1-4}\end{tabular}$

Let start Balancing the equation,

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Step-1

Now, We will first balance Carbon (C)

As there is 1 C in L.H.S (Initial) and 1 C in R.H.S (Final), we can see that the carbon (C) atom is already balanced as L.H.S = R.H.S

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}C&1&1\\C&1&1\\C&1&1\end{tabular}$

Therefore, Carbon (C) is balanced.

Hence the first step equation will be, same as that of the given equation,

$\displaystyle\sf CH_{4} + O_{2 }\longrightarrow CO_{2} + H_{2}O+Heat$

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Step-2

Secondly, Balancing the Hydrogen atom.

As there is 4 H in L.H.S (Initial) and 2 H in R.H.S (Final), Multiplying with the common factor (2) on the Final side, i.e. 2 × 2 = 4 ; gives 4 H atoms in R.H.S (Final) which equals L.H.S (Initial).

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}H&4&2\\H&4&2\times2\\H&4&4\end{tabular}$

Therefore, Hydrogen (H) is balanced.

Hence, the equation becomes,

$\displaystyle\sf CH_{4} + O_{2 }\longrightarrow CO_{2} + 2\;H_{2}O+Heat$

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Step-3

Now,balancing Oxygen (O)

On the R.H.S the Oxygen (2 H₂O) will be => 2 × 1 = 2 and 2 from CO₂ i.e. total will be 2 + 2 = 4.

As there is 2 O in L.H.S (Initial) and 4 O in R.H.S (Final), Multiplying with the common factor (2) on the Initial side, i.e. 2 × 2 = 4 ; gives 4 O atoms in L.H.S (Initial) which equals R.H.S (Final).

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}O&2&4\\O&2\times2&4\\O&4&4\end{tabular}$

Therefore, Oxygen (O) is balanced.

Equation becomes,

$\displaystyle\sf CH_{4} + 2\;O_{2 }\longrightarrow CO_{2} + 2\;H_{2}O+Heat$

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Now, Checking the concentration of each atom again in the Step-3 step.

$\large\begin{tabular}{|c|c|c|c|}\cline{1-4}Element&C&O&H\\\cline{1-4} Initial&1&4&4\\\cline{1-4}Final&1&4&4\\\cline{1-4}\end{tabular}$

Here, we can see, as all atoms are balanced the third step equation is a balanced equation of the given Equation.

Therefore,

$\displaystyle\large{\underline{\boxed{\red{\sf CH_{4} + 2\;O_{2 }\longrightarrow CO_{2} + 2\;H_{2}O+Heat}}}}$

Hence, the equation is balanced !

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