Balancing Equations
The first term is the reactants. Reactants are the substances on the left side of an equation, before the arrow.
Products are the terms on the right side of the equation, after the arrow. Chemistry Balancing Equations
Logically these two terms make sense. The products are what you have before the reaction, and the reactants are what you have after the reaction has occurred.
In order to have a balanced equation, both the reactants and products must have the same number of atoms.
When this has been achieved, the equation is said to have been balanced. The below equation is balanced since the reactants equal the products. We have 4 H’s on the left side and 4 H’s on the right side, 2 O’s on the left side and 2 O’s on the right side.
Requirements to balance an equation
Good question! In order to balance an equation, you can only modify the coefficients (the numbers in front of the atom). Chemistry Balancing Equations
For example, I can only modify the 2 in front of the H_{2 }(on the reactants side), I cannot modify the subscript on the H and make that an H_{1}. This would make the world implode, and we don’t want that to happen.
Therefore, we tweak with the coefficients on each of the terms. Another Error that is frequently made is trying to add coefficients where they don’t belong.
On the product side of the above equation, I cannot stick in a coefficient on the O. What I’m trying to say is, you cannot make the above equation balanced by doing this:
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Therefore, you can only modify the coefficients as a term. Let’s see if you get the hang of it with a simple balancing problem.
Problem 1 |
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Problem: Which are reactants and which are products in the following reaction?2H_{2}O_{(l) }2H_{2}O(g) Plan: By looking at the definitions above, the reactants are on the left side of the equation and the products are on the right side of the equation. Chemistry Balancing Equations |
Problem 2 |
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Problem: Balance the following equation N_{2(g)}+ H_{2(g) }NH_{3(g)} Plan: We see that we have 2 N’s on the reactants side and 1 N on the product side, therefore we must multiply the product side by 2. |
Now that we have the general idea behind balancing, let’s try a harder problem.
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Problem 3 |
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Problem: Balance the following equation C_{6}H_{12}O_{6(s)} + O_{2(g)} CO_{2(g)} + H_{2}O_{(l)} Plan: The first step is to put a coefficient of 1 in front of the longest term. This helps you see what needs to be done to make the complicated term equal to the non-complicated terms. C_{6}H_{12}O_{6} is the longest term, therefore a 1 is placed in front of it. |
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