Question

Balance the equation given below :-

$\sf {Fe\ +\ H_2O \to Fe_3O_4\ +\ H_2}$

Answer 1

Step-by-step explanation:

$\Large{\underline{\underline{\sf{\purple{Required\ Answer:-}}}}}$

✯ Here we are given a chemical equation and are asked to balance it. The equation is as follows :-

$\sf {Fe\ +\ H_2O \to Fe_3O_4\ +\ H_2}$

$\: \: \: \:$ To balance the above equation we have to follow some steps and will get our final balanced equation.

$\bf {Step\ I:}$ To balance a chemical equation, first draw boxes around each formula. Do not change anything inside the boxes while balancing the equation.

$\: \: \: \: \boxed {\sf{Fe}}\ +\ \boxed {\sf{H_2O}} \to \boxed {\sf{Fe_3O_4}}\ +\ \boxed {\sf{H_2}}$

$\bf {Step\ II:}$ List the number of atoms of different elements present in the unbalanced equation in Step 1.

$\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c |c} \tt{Element}& \tt{Number\ of\ atoms\ in\ reactants\ (L.H.S)}& \tt{Number\ of\ atoms\ in\ products\ (R.H.S)} \\ \dfrac {\qquad\qquad}{ \sf Fe}&\dfrac{\qquad\qquad}{ \sf 1}&\dfrac{\qquad\qquad}{ \sf 3}& \\ \dfrac{\qquad\qquad}{ \sf H}&\dfrac{\qquad\qquad}{ \sf 2}&\dfrac{\qquad\qquad}{ \sf 2}& \\ \dfrac{\qquad\qquad}{ \sf O}&\dfrac{\qquad\qquad}{ \sf 1}&\dfrac{\qquad\qquad}{ \sf 4}&\end{array}} \end{gathered}\end{gathered}\end{gathered}$

$\bf {Step\ III:}$ It is often convenient to start balancing with the compound that contains the maximum number of atoms. It may be a reactant or a product. In that compound, select the element which has the maximum number of atoms. Using these criteria, we select $\sf {Fe_3O_4}$ and the element oxygen in it. There are four oxygen atoms on the R.H.S and only one on the L.H.S.

$\: \: \: \:$ To balance the oxygen atoms :-

$\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c |c} \tt{Atoms\ of\ oxygen}& \tt{In\ reactants}& \tt{In\ products} \\ \dfrac{\qquad\qquad}{ \sf (i)\ Initial}&\dfrac{\qquad\qquad}{ \sf 1\ (in\ H_2O)}&\dfrac{\qquad\qquad}{ \sf 4\ (in\ Fe_3O_4)}& \\ \dfrac{\qquad\qquad}{ \sf (ii)\ To\ balance}&\dfrac{\qquad\qquad}{ \sf 1 \times 4}&\dfrac{\qquad\qquad}{ \sf 4}&\end{array}} \end{gathered}\end{gathered}\end{gathered}$

$\: \: \: \:$ To equalise the no. of atoms, it must be remembered that we cannot alter the formulae of the compounds or elements involved in the reactions. For example, to balance oxygen atoms we can put coefficient '4' as 4 $\sf {H_2O}$ and not $\sf {H_2O_4}$ or $\sf {(H_2O)_4}$ . Now the equation becomes :-

$\: \: \: \: {\underbrace{\underline{\boxed{\sf Fe}\ +\ 4 \boxed{\sf H_2O} \to \boxed{\sf Fe_3O_4}\ +\ \boxed{\sf H_2}}}_{\tiny \rm {\blue{Partly\ Balanced\ Equation}}}}$

$\bf {Step\ IV:}$ Fe and H atoms are still not balanced. Pick any of these elements to proceed further. Let us balance hydrogen atoms in the partly balanced equation.

$\: \: \: \:$ To equalise the no. of H atoms, make the no. of molecules of hydrogen as four on the R.H.S.

$\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c |c} \tt{Atoms\ of\ hydrogen}& \tt{In\ reactants}& \tt{In\ products} \\ \dfrac{\qquad\qquad}{ \sf (i)\ Initial}&\dfrac{\qquad\qquad}{ \sf 8\ (in\ 4\ H_2O)}&\dfrac{\qquad\qquad}{ \sf 2\ (in\ H_2)}& \\ \dfrac{\qquad\qquad}{ \sf (ii)\ To\ balance}&\dfrac{\qquad\qquad}{ \sf 8}&\dfrac{\qquad\qquad}{ \sf 2 \times 4}&\end{array}} \end{gathered}\end{gathered}\end{gathered}$

$\: \: \: \:$ The equation would be :-

$\: \: \: \: {\underbrace{\underline{\boxed{\sf Fe}\ +\ 4\ \boxed{\sf H_2O} \to \boxed{\sf Fe_3O_4}\ +\ 4 \boxed{\sf H_2}}}_{\tiny \rm {\blue{Partly\ Balanced\ Equation}}}}$

$\bf {Step\ V:}$ Examine the above equation and pick up the third element which is not balanced. We find that only one element is left to be balanced, that is, iron.

$\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c |c |c} \tt{Atoms\ of\ iron}& \tt{In\ reactants}& \tt{In\ products} \\ \dfrac{\qquad\qquad}{ \sf (i)\ Initial}&\dfrac{\qquad\qquad}{ \sf 1\ (in\ Fe)}&\dfrac{\qquad\qquad}{ \sf 3\ (in\ Fe_3O_4)}& \\ \dfrac{\qquad\qquad}{ \sf (ii)\ To\ balance}&\dfrac{\qquad\qquad}{ \sf 1 \times 3}&\dfrac{\qquad\qquad}{ \sf 3}&\end{array}} \end{gathered}\end{gathered}\end{gathered}$

$\: \: \: \:$ To equalise Fe, we take three atoms of Fe on the L.H.S.

$\: \: \: \: \sf 3\ \boxed {\sf Fe}\ +\ \sf 4\ \boxed {\sf H_2O} \to \boxed {\sf Fe_3O_4}\ +\ \sf 4\ \boxed {\sf H_2}$

$\bf {Step\ VI:}$ Finally, to check the correctness of the balanced equation, we count atoms of each element on both sides of the equation.

$\: \: \: \: {\underbrace{\underline{\sf {3Fe\ +\ 4H_2O \to Fe_3O_4\ +\ 4H_2}}}_{\tiny \rm {\pink{Required\ Balanced\ Equation}}}}$

Now the chemical equation is balanced as the no. of atoms of elements on both sides of equation are equal.

Answer 2

Step-by-step explanation:

Hope it HELPS you....!!!!