Question

How would you balance the reaction of white phosphorus $({P_4})$ with oxygen $({O_2})$ which produces phosphorus oxide $({P_4}{O_{10}})$ ?

Answer 1

See the number of atoms on both sides of the reaction. If they are not the same , try to make the coefficient of each reactant or product change until you get them balanced. White phosphorus is written as ${P_4}$ and thus we have multiples of four to assign in place of coefficient and then similar numeric changes are made to the other side of reaction to balance phosphorus.

Complete step by step answer:
We have three molecules here which react in a chemical reaction. The reaction of two molecules gives the third product. The unbalanced chemical reaction is
${P_4} + {O_2} \to {P_4}{O_{10}}$
Here we have $4P$on the left side of reaction and similarly $4P$ as in molecule ${P_4}{O_{10}}$ .
So now we can say just by seeing the above equation that phosphorus atoms are balanced but now the other part of the reaction is to balance oxygen atoms.
To balance oxygen atoms, firstly see how many atoms are there in each side of the reaction. We have $2O$ on the left hand side as ${O_2}$ and $10\,oxygen$ on the right hand side of the equation as ${P_4}{O_{10}}$ .
To balance them or to make them equal we have two choices, first is to multiply ${P_4}{O_{10}}$ by some number and the second choice is that we can multiply ${O_2}$ by some number. There are respected effects of each choice. If we multiply ${P_4}{O_{10}}$ it will unbalance the phosphorus atoms so we cannot attempt this step of multiplication. Now our second step is to multiply ${O_2}$ atom by some number and we will see that after the multiplication the other molecules don't get unbalanced.
So, when we multiply ${O_2}$ by $5$, it will form a total of $10\,\,oxygen$ on the left hand side of the reaction. Now why we multiply by $5$ and not by any other number, is because we have to make the left hand side and right hand side reaction the same or equal.
If we now write the balance equation it will be like this,
${P_4} + \,5{O_2} \to {P_4}{O_{10}}$
Here, let’s count the number of atoms of each type. We have $4P,\,10\,O$on left hand side as ${P_4}$ and $5{O_2}$ while on right hand side equation, we have same number of atoms as ${P_4}{O_{10}}$ .

Note: Don’t multiply the right hand side molecule of ${P_4}{O_{10}}$ , because after multiplication the number of atoms increase on its side, we have to balance them so try to multiply the smaller terms. If you are balancing the chemical equation for the first time, then try to multiply the coefficient as you can by any number, it will take some time but at last you will get a balanced chemical equation.