How to balance complicated chemical equations?
Answer
581.1k+ views
Hint: In this question, we will learn how to balance the chemical equation and why it is important to balance them. The chemical equation should be balanced because The number of atoms on both sides of the equation should be equal as it sticks to the Law of conservation.
Complete answer:
In this, we will learn the method to balance a chemical equation. Balancing a chemical equation can sometimes be a bit of a task.
Let us consider an Equation for a better understanding-
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,C{{O}_{2}}\,+\,{{H}_{2}}O$
The first thing we do is start balancing the carbon in the equation.
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,{{H}_{2}}O$
Now, we can see that the Carbon is balanced. The next thing to do is to balance the hydrogen atom. Here we have six Hydrogens on the reactant side and only two hydrogens on the product side. So, we have to balance it.
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,3{{H}_{2}}O$
We have balanced the hydrogen but The Oxygen atom is still unbalanced.
Now the next step is to balance the oxygen. There are two oxygen on the reactant side and seven oxygen on the product side. To add a stoichiometric coefficient on the reactant side to balance the oxygen on the product side is kind of complicated but we can go with fraction.
${{C}_{2}}{{H}_{6\,}}\,+\,\dfrac{7}{2}{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,3{{H}_{2}}O$
Look at the equation now –
Oxygens on the reactant side are ($\dfrac{7}{2}\times 2$) = 7 which is the same as on the product side.
But the issue is In the Final balanced chemical equation, the fraction does not look appealing. One should eliminate the fraction. We can multiply the entire chemical equation with a number but the only condition is we have to do it on both sides of the reaction.
We can multiply both sides with 2 to eliminate the fraction
$2{{C}_{2}}{{H}_{6\,}}\,+\,7{{O}_{2\,}}\,\to \,\,4C{{O}_{2}}\,+\,6{{H}_{2}}O$
Now the chemical equation is properly balanced.
Note:
Look for the most complicated compound. Try to add a stoichiometric coefficient on both the sides. If there are any polyatomic ions on both sides of the equation, try to balance them. Leave the simpler molecules for the end. If there is any fractional coefficient, multiply both the reactant and product side with the number in the denominator. At last, see the numbers of atoms on both the sides of the equation are equal to make sure that your chemical equation is properly balanced.
Complete answer:
In this, we will learn the method to balance a chemical equation. Balancing a chemical equation can sometimes be a bit of a task.
Let us consider an Equation for a better understanding-
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,C{{O}_{2}}\,+\,{{H}_{2}}O$
The first thing we do is start balancing the carbon in the equation.
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,{{H}_{2}}O$
Now, we can see that the Carbon is balanced. The next thing to do is to balance the hydrogen atom. Here we have six Hydrogens on the reactant side and only two hydrogens on the product side. So, we have to balance it.
${{C}_{2}}{{H}_{6\,}}\,+\,{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,3{{H}_{2}}O$
We have balanced the hydrogen but The Oxygen atom is still unbalanced.
Now the next step is to balance the oxygen. There are two oxygen on the reactant side and seven oxygen on the product side. To add a stoichiometric coefficient on the reactant side to balance the oxygen on the product side is kind of complicated but we can go with fraction.
${{C}_{2}}{{H}_{6\,}}\,+\,\dfrac{7}{2}{{O}_{2\,}}\,\to \,\,2C{{O}_{2}}\,+\,3{{H}_{2}}O$
Look at the equation now –
Oxygens on the reactant side are ($\dfrac{7}{2}\times 2$) = 7 which is the same as on the product side.
But the issue is In the Final balanced chemical equation, the fraction does not look appealing. One should eliminate the fraction. We can multiply the entire chemical equation with a number but the only condition is we have to do it on both sides of the reaction.
We can multiply both sides with 2 to eliminate the fraction
$2{{C}_{2}}{{H}_{6\,}}\,+\,7{{O}_{2\,}}\,\to \,\,4C{{O}_{2}}\,+\,6{{H}_{2}}O$
Now the chemical equation is properly balanced.
Note:
Look for the most complicated compound. Try to add a stoichiometric coefficient on both the sides. If there are any polyatomic ions on both sides of the equation, try to balance them. Leave the simpler molecules for the end. If there is any fractional coefficient, multiply both the reactant and product side with the number in the denominator. At last, see the numbers of atoms on both the sides of the equation are equal to make sure that your chemical equation is properly balanced.
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