What are some efficient approaches for balancing long chemical equations? [closed]








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I am having some difficulties in balancing chemical equations involving many elements. It takes a lot of time (for me) to balance the coefficients **using the hit and trial method**. Especially like this one:
$$\ce{K2Cr2O7 + HCl -> KCl + CrCl3 + H2O + Cl2}$$


**My problem is:** Where should I start? The heavier or the lighter elements? Which is more efficient?


For the above problem, I started with K since it appears first in the equation. I don't think this is a great way for approaching these questions.




Let's start by the atoms of the most complicated molecule : $\ce{K2Cr2O7}$.


Start with $\ce{K}$. There is $2$ $\ce{K}$ on the left hand side. They must appear in the products : $2$ $\ce{KCl}$. We'll take care of the $\ce{Cl}$ atoms later on. Let's finish the starting molecule : $\ce{K2Cr2O7}$


There is also $2$ $\ce{Cr}$ atoms in the reagents. So $2$ $\ce{Cr}$ atom must appear in the products : $2$ $ \ce{CrCl3}$.


And now there are $7$ $\ce{O}$ atoms in $\ce{K2Cr2O7}$. As a consequence, $7$ $\ce{H2O}$ molecules must appear in the products, with $14$ $\ce{H}$ atoms, and of course $14$ $\ce{HCl}$ molecules in the reagents.


Now the first molecule of the reagents is OK. Let's consider what atoms have been introduced in the products which are NOT YET equilibrated, namely the chlorine atoms.


In total, $8$ $\ce{Cl}$ atoms are included in $2 (\ce{KCl + CrCl3}$). These $8$ chlorine atoms must appear in the reagents : It makes $8$ $\ce{HCl}$ molecules in the reagents. We have already seen that $14$ $\ce{HCl}$ has been produced, and here only $8$ chlorine atoms are needed for producing $2$ $\ce{KCl}$ and $2$ $\ce{CrCl3}$. The $14 - 8 = 6$ chlorine atoms in excess are used to produce the $\ce{Cl2}$ molecules. And to produce exactly $3$ $\ce{Cl2}$ molecules (with $6$ $\ce{Cl}$). So the final equation is :$$\ce{K2Cr2O7 + 14 HCl -> 2 KCl + 2 CrCl3 + 7 H2O + 3 Cl2}$$ This result can also be obtained with the half-equation method, as suggested in Mithoron's answer.


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