Question

Balance the equation:
 $ ZnS + {O_2} \to ZnO + S{O_2} $

Answer 1

Hint: In order to balance a chemical reaction, we have to make sure that the number of each element on the reactant side must be equal to that of the product side. In order to ensure the law of conservation of matter a chemical equation should be balanced.

Complete step by step solution:
In order to ensure a given reaction is balanced the number of elements on the left-hand side (lhs) must be equal to that of the number of elements on the right-hand side (rhs).
Consider the given chemical equation:
 $ ZnS + {O_2} \to ZnO + S{O_2} $
Consider the number of moles of each element on both reactant and product side:
On the LHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 1 \\
  S - 1 \\
  O - 2 \\
\end{gathered} $
On the RHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 1 \\
  S - 1 \\
  O - 3 \\
\end{gathered} $
Since the number of moles of oxygen on the LHS and RHS are not the same , we have to balance it.
In order to balance the number of moles of oxygen, make the coefficient of $ {O_2} - 2 $ and the coefficient of $ ZnO - 2 $ ; Then the equation will be:
 $ ZnS + 2{O_2} \to 2ZnO + S{O_2} $
On the LHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 1 \\
  S - 1 \\
  O - 4 \\
\end{gathered} $
On the RHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 1 \\
  O - 4 \\
\end{gathered} $
Now the number of moles of $ Zn $ on the LHS and RHS are different.
In order to balance their number, make the coefficient of $ ZnS - 2 $ ;Then the equation will be:
 $ 2ZnS + 2{O_2} \to 2ZnO + S{O_2} $
On the LHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 2 \\
  O - 4 \\
\end{gathered} $
On the RHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 1 \\
  O - 4 \\
\end{gathered} $
Here the number of moles of $ S $ on the LHS and RHS are different.
In order to balance their number, make the coefficient of $ S{O_2} - 2 $ ;Then the equation will be:
 $ 2ZnS + 2{O_2} \to 2ZnO + 2S{O_2} $
On the LHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 2 \\
  O - 4 \\
\end{gathered} $
On the RHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 2 \\
  O - 6 \\
\end{gathered} $
Here the number of moles of $ O $ on the LHS and RHS are different.
In order to balance their number, make the coefficient of $ {O_2} - 3 $ ;Then the equation will be:
 $ 2ZnS + 3{O_2} \to 2ZnO + 2S{O_2} $
On the LHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 2 \\
  O - 6 \\
\end{gathered} $
On the RHS (no of moles of each element):
 $ \begin{gathered}
  Zn - 2 \\
  S - 2 \\
  O - 6 \\
\end{gathered} $
Now the number of moles of each element on both sides has been equal and we got the balance chemical equation as:
 $ 2ZnS + 3{O_2} \to 2ZnO + 2S{O_2} $

Note:
While balancing the chemical reactions by making the number of atoms same on both sides we also have to ensure the net charge is also same on both reactant and product side. Also, while balancing we are not supposed to alter the chemical formula at any cause.