Who tells me this reaction?

Question

$$P_4O_{10} + H_2O\rightarrow H_3PO_4$$ I've a solution what I couldn't sense. How did we find this matrix and vectors? as you can see: $$\begin{bmatrix} 4 & 1 & 0 \\ 10 & 4 & 1 \\ 0 & 3 & 2 \\ \end{bmatrix}$$ If someone tells me how did we find vector of a reaction?

Answer 1

What you want to do is to balance the equation between its constitutive elements.

You are searching $x,y,z$ such that $x\ P_4O_{10}+y\ H_2O\to z\ H_3PO_4$

$\begin{cases} \text{balancing P}:4x=z\\ \text{balancing O}:10x+y=4z\\ \text{balancing H}:2y=3z \end{cases}\iff\begin{cases} 4x+0y=1z\\ 10x+1y=4z\\ 0x+2y=3z \end{cases}\iff\begin{pmatrix} 4 & 0 \\ 10 & 1 \\ 0 & 2\end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} 1 \\ 4 \\ 3\end{pmatrix} z$

But you could as well express it like this, putting $z$ on left side with a minus sign.

$\begin{pmatrix} 4 & 0 & -1 \\ 10 & 1 & -4 \\ 0 & 2 & -3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}=0$

Now look at this presentation to see how it is built $\begin{vmatrix} P_4 & P_0 &\to & P_1\\ O_{10} & O_1&\to & O_4\\ H_0 & H_2 &\to &H_3 \end{vmatrix}$

This matrix presentation is generally useful for equations with many products involved, although with simple reactions like this one, it is generally manageable by hand from the first system :

$\begin{cases} 4x=z\\ 10x+y=4z\\ 2y=3z \end{cases}\iff \begin{cases} z=4x\\ 10x+6x=16x\\ y=6x \end{cases}\iff \begin{cases} z=4x\\ y=6x \end{cases}$

Since $x,y,z$ are integer in chemical reactions, $x=1,y=6,z=4$ works well.

$P_4O_{10}+6\,H_2O\to 4\,H_3PO_4$

Answer 2

I would write the matrix differently. You have three variables (number of molecules of each of the compounds) and three equations (stating that the numbers of atoms of each element must balance). For example, if there are $a$ molecules of $P_4O_{10}$, $b$ molecules of $H_2O$, and $c$ molecules of $H_3PO_4$, saying that the number of phosphorus atoms must balance would be the equation $4a+0b-c=0$. I would end up with the matrix $$\begin{bmatrix} 4 & 0 & -1 \\ 10 & 1 & -4 \\ 0 & 2 & -3 \end{bmatrix}.$$

Answer 3

Why do it via a matrix at all: maybe simple number theory can help us. If the three amounts are called $a,b,c$ the amount of $P$'s tells us that $c= 4a$ while looking at the $H$'s we get $2b = 3c$. Then finally the $O$'s tell us that $10a+ b = 4c$,

The gcm of the numbers involved is $\gcd(4,3,2,10) = 60$, so we multiply the equation $10a + b = 4c$ by 6, to make all the coefficients multiples of the gcm and we get $60a + 6b = 24c$, or using the other relations $4a=c, 2b=3c$ we get $15c + 9c = 24c$, which comes out nicely. So $(a,b,c) =(15, 9, 24)$ works but we can reduce that to $(5,3,8)$ dividing by $3$. So we have a (minimal) solution.