Question

Fe + H₂0 gives
Fe3O4 + H₂......balance this equation !??​

Answer 1

$Skeletal:\\Fe+H_2O---->Fe_3O_4+H_2\\\\Balanced:\\3Fe+4H_2O---->Fe_3O_4+4H_2$

Answer 2

Answer:

$\displaystyle\sf{3\;Fe\;+\;4\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;4\;H_{2}}$

Explanation:

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The given Reaction is ,

$\displaystyle\sf{Fe\;+\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;H_{2}}$

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

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

Let start Balancing the equation,

Step-1

Now, We will first balance Iron (Fe)

As there is 1 Fe in L.H.S (Initial) and 3 Fe in R.H.S (Final), Multiplying with the common factor (3) on the initial side, i.e. 1 × 3 = 3 ; gives 3 Fe atoms in L.H.S (Initial) which equals R.H.S (Final).

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}Fe&1&3\\Fe&1\times3&3\\Fe&3&3\end{tabular}$

Therefore, Iron (Fe) is balanced.

Hence the first step equation will be,

$\displaystyle\sf{3\;Fe\;+\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;H_{2}}$

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Step-2

Now, We will balance oxygen (O)

As there is 1 O in L.H.S (Initial) and 4 O in R.H.S (Final), Multiplying with the common factor (4) on the initial side, i.e. 1 × 4 = 4 ; gives 4 O atoms in L.H.S (Initial) which equals R.H.S (Final).

(The number 4 will be placed against the H₂O molecule)

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}O&1&4\\O&1\times4&4\\O&4&4\end{tabular}$

Therefore, Oxygen (O) is balanced.

Hence the second step equation will be,

$\displaystyle\sf{3\;Fe\;+\;4\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;H_{2}}$

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Step-3

Now, We will balance Hydrogen (H)

On the L.H.S the hydrogen (4 H₂O) will be => 4 × 2 = 8

As there is (4×2=8) 8 H in L.H.S (Initial) and 2 H in R.H.S (Final), Multiplying with the common factor (4) on the Final side, i.e. 2 × 4 = 8 ; gives 8 H atoms in R.H.S (Final) which equals L.H.S (Initial).

$\large\begin{tabular}{c|c|c}Element.&L.H.S&R.H.S\\\cline{1-3}H&8&2\\H&8&2\times4\\H&8&8\end{tabular}$

Therefore, Hydrogen (H) is balanced.

Hence the Third step equation will be,

$\displaystyle\sf{3\;Fe\;+\;4\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;4\;H_{2}}$

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Now, comparing the L.H.S & R.H.S in the third step equation, will be

$\begin{tabular}{|c|c|c|c|}\cline{1-4}Element&Fe&O&H\\\cline{1-4} Initial&3&4&8\\\cline{1-4}Final&3&4&8\\\cline{1-4}\end{tabular}$

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

Therefore, the Balanced equation will be,

$\displaystyle\bold{\red{3\;Fe\;+\;4\;H_{2}O \longrightarrow Fe_{\;3}O_{\;4}\;+\;4\;H_{2}}}$

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