Question

Balance the chemical equation Mg+O2=MgO

Answer 1

Answer:

Explanation:Mg2 +O2=2MgO

Answer 2

Answer:

$\sf 2\;Mg+O_{2}\longrightarrow 2\;Mg\;O$

Explanation:

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The given Reaction is,

$\displaystyle\sf{Mg+O_{2}\longrightarrow Mg\;O}$

Now, Checking the initial & final concentration of both elements.

$\begin{tabular}{|c|c|c|}\cline{1-3}Element&Mg&O\\\cline{1-3} Initial&1&2\\\cline{1-3}Final&1&1\\\cline{1-3}\end{tabular}$

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Now Balancing the equation,

Step-1

Now, We will first balance Oxygen (O)

As there is 2 O in L.H.S (Initial) and 1 O in R.H.S (Final), Multiplying with the common factor (2) on the Final (R.H.S) side, i.e. 1 × 2 = 2 ; gives 2 Oxygen atoms in R.H.S (Final) which equals L.H.S (Initial).

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

Therefore, Oxygen (O) is balanced.

Hence the first step equation will be,

$\displaystyle\sf{Mg+O_{2}\longrightarrow 2\;Mg\;O}$

Here 2 is written in front of that compound as we know we cannot violate formula of compound to balance the equation.

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Step-2

Now, Balancing Magnesium (Mg) atom.

On the R.H.S the Magnesium (2 Mg O) will be 1 × 2 = 2

As there is (1×2=2) 2 Mg in R.H.S (Final) and 1 Mg in L.H.S (Initial), Multiplying with the common factor (2) on the Initial side, i.e. 2 × 1 = 2 ; gives 2 Mg atoms in L.H.S (Initial) which equals R.H.S (Final).

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

Therefore, Magnesium (Mg) is balanced.

Therefore, the second step equation will be

$\sf 2\;Mg+O_{2}\longrightarrow 2\;Mg\;O$

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As we have tried to balanced both the atoms, lets check their initial and Final Amounts,

$\begin{tabular}{|c|c|c|}\cline{1-3}Element&Mg&O\\\cline{1-3} Initial&2&2\\\cline{1-3}Final&2&2\\\cline{1-3}\end{tabular}$

As we can see the atoms are balanced, we get a balanced equation,

So, the Balanced equation of the given equation is

$\red{\bf 2\;Mg+O_{2}\longrightarrow 2\;Mg\;O}$

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