Balancing reactions algebraic method limitations

Question

How do I balance this reaction using the algebraic method?

Seems that the equation can't be solved.

How do I balance such reaction and does the algebraic method has limitations and wont work for balancing some reactions?

$$\ce{XeF4 + H2O -> XeO3 + Xe + O2 + HF}$$

$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF}$$

$$4 a = f$$

$$2 b = f$$

$$a = c + d$$

$$ b = 3 c + 2 e $$

Answer 1

As per the OP's request, here is a solution. The equation is:

$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$

So atom balance yields:

$$4a=1f \tag 2$$ $$2b=1f \tag 3$$ $$1a=1c + 1d \tag 4$$ $$1b = 3c + 2e \tag 5$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

1. Assume $a = 1$, which is the minimum non-zero value it might have. Then, from equation $(4)$, $1 = a = c + d$, which is impossible because either $c$ or $d$ would have to be zero.

2. Assume $a = 2$. Then $b = 4$ because equations $(2)$ and $(3)$ show that $b = 2a$. But then equation $(5)$ shows that $4 = b = 3c + 2e$. Then if $c = 1, e = 0.5$, which is impossible. Obviously, $c$ cannot equal $2$, because that would give a negative value for $e$.

3. Assume $a = 3$. Then $b = 6$, because $b = 2a$, as above. Then equation $(5)$ shows that $6 = b = 3c + 2e$. Then if $c = 1, e = 1.5$, which is impossible. If $c = 2$, then $e = 0$, which is impossible. As before, $c$ cannot be greater than $2$, because that would give a negative value for $e$.

4. Assume $a = 4$. Then $b = 8$ and $f = 2b = 16$. Then equation $(5)$ shows that $8 = b = 3c + 2e$. If $c = 1$, then $2e = 5$, so $e = 2.5$, which is impossible. If $c = 2$, then $2e = 2$, so $e = 1$. Then equation $(4)$ shows that $4 = 2 + d$, so $d = 2$. This is a solution. Lastly, $c$ cannot equal $3$, because that would give a negative value for $e$.

Almost final result: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence

$$\ce{4 XeF4 + 8 H2O -> 2 XeO3 + 2 Xe + 1 O2 + 16 HF \tag 6}$$

For a = 5, the solution is a = 5, b = 10, c = 2, d = 3, e = 2 and f = 20. This is the same as Oscar Lanzi's example solution. For a = 6, the solution is a = 6, b = 12, c = 2, d = 4, e = 3 and f = 24.

For a = 7, a double solution occurs: a = 7, b = 14, c = 2, d = 5, e = 4 and f = 28 and also a = 7, b = 14, c = 4, d = 3, e = 1 and f = 28.

There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is 11a.

Bonus: Mithoron provided a link to a problem Balancing disproportionation reactions answered by Ivan Neretin years ago. The equation that was to be balanced was:

$$\ce{a XeF2 + b H2O -> c Xe + d XeO3 + e O2 + f HF \tag 7}$$

So atom balance yields:

$$1a=1c + 1d \tag 8$$ $$2a=1f \tag 9$$ $$2b=1f \tag {10}$$ $$1b = 3d + 2e \tag {11}$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I first note that equations $(9)$ and $(10)$ show that $a = b$. Then equation $(11)$ shows that $b$ must at least equal $5$. Otherwise either $d = 0$ or $e = 0$, which is impossible.

Now I proceed by assuming $a = b = 5$. Then equation $(8)$ gives rise to the following four possibilities: $c = 1$ and $d = 4$; $c = 2$ and $d = 3$; $c = 3$ and $d = 2$; $c = 4$ and $d = 1$. If $c = 1$ and $d = 4$, then equation $(11)$ yields $5 = 12 + 2e$, which is impossible. Likewise for $c = 2$ and $d = 3$ and also for $c = 3$ and $d = 2$. But if $c = 4$ and $d = 1$, then equation $(11)$ yields $5 = 3 + 2e$, so $e = 1$. This is a solution: $a = 5, b = 5, c = 4, d = 1, e = 1,$ and $f = 10$. Hence the balanced equation is:

$$\ce{5 XeF2 + 5 H2O -> 4 Xe + 1 XeO3 + 1 O2 + 10 HF }$$

However, as Ivan Neretin beautifully pointed out in his answer, there is also the simpler redox reaction given by

$$\ce{2 XeF2 + 2 H2O -> 2 Xe + 1 O2 + 4 HF }$$

So any linear combination of these, with positive integer coefficients, is also a valid balanced equation. For example, just add the last two equations together together:

$$\ce{7 XeF2 + 7 H2O -> 6 Xe + 1 XeO3 + 2 O2 + 14 HF }$$

Answer 2

Actually there is no unique solution. You have four linearly independent equations and six compounds. In a typical redox reaction you need one less equation than the number of compounds. When the difference is greater than one, you have a combination of multiple reactions.

We can get unique numbers only by assuming that just one oxidation product is formed, either $\ce{XeO3}$ or $\ce{O2}$.

If we choose $\ce{XeO3}$, then we can use the redox method to see that two moles of that compound are formed per one mole of elemental xenon. This leads to the result

$\ce{3 XeF4 + 6 H2O -> Xe + 2 XeO3 + 12 HF}$

If instead $\ce{O2}$ is chosen, one mole of that matches with one mole of xenon being formed:

$\ce{XeF4 + 2 H2O -> Xe + O2 + 4 HF}$

Now any linear combination you want is a balanced reaction, corresponding to some assumed ratio of $\ce{XeO3}$ versus $\ce{O2}$ formation. If you assume equal molar amounts of both products, for instance, you would have the first component plus twice the second:

$\ce{5 XeF4 + 10 H2O -> 3 Xe + 2 XeO3 + 2 O2 + 20 HF}$

A very different answer from others yet equally well balanced!

In a situation like this you have to measure the relative contributions of the two oxidation products to get the true stoichiometry. And specify the conditions of such an experiment, for no law of nature requires the ratio to remain constant.

Answer 3

I like how Ed V has explained the mathematical solution logically. I'd like to show another way of solving the problem. The equation is as written by OP:

$$\ce{a XeF4 + b H2O -> c XeO3 + d Xe + e O2 + f HF \tag 1}$$

So atom balance of equation $(1)$ yields:

$$\text{Balancing } \ce{F}: \quad 4a = f \tag 2$$ $$\text{Balancing } \ce{H}: \quad 2b = f \tag 3$$ $$\text{Balancing } \ce{Xe}: \quad a = c + d \tag 4$$ $$\text{Balancing } \ce{O}: \quad b = 3c + 2e \tag 5$$

The four equations relate to the six unknown coefficients $a$ through $f$. As Ed V pointed out, each coefficient is an integer greater than zero, so I proceed as follows:

1. I assumed $a = 2$, because in right hand side there are two $\ce{Xe}$ atoms. Then, from equation $(2)$, $f = 8$ and from equation $(3)$, $f = 8 = 2b \Rightarrow \; \therefore b = 4$.

2. Because coefficients are non-zero integrals, I also assumed $c = d= \frac12 a =\frac12 \times 2 = 1$. Thus, from equation $(5)$, $2e = b - 3c = 4 - 3 = 1 \Rightarrow \; \therefore b = \frac12$. This is not an integer so my original assumption of $a = 2$ is incorrect. Yet, we can complete the task at this point because fraction of coefficients are allowed in chemistry if at least one coefficient is an integral. Therefore, the equation would be:

$$\ce{2 XeF4 + 4 H2O -> 1 XeO3 + 1 Xe + 1/2 O2 + 8 HF \tag 6}$$

If you multiply whole equation $(6)$ by 2, it becomes:

$$\ce{4 XeF4 + 8 H2O -> 2 XeO3 + 2 Xe + 1 O2 + 16 HF \tag 7}$$

The equation $(7)$ is the correct answer, identical to Ed V's. If this is not acceptable, then we can continue as follows:

3. Since we get $e = \frac12$ by assuming $a=2$, we can start again assuming $a = 4$ as stultification of 2. Then, from equation $(2)$, $f = 16$ and from equation $(3)$, $f = 16 = 2b \Rightarrow \; \therefore b = 8$.

4. Apply this values to equation $(4)$ and $(5)$:

$$4 = c + d \tag 8$$ $$8 = 3c + 2e \tag 9$$

Subtract equation $(9)$ from equation $(8) \times 3$:

$$3d - 2e = 4 \tag 10$$

The smallest values of $d$ and $e$ to fulfill this requirement are $d = 2$ and $e = 1$. Hence, from the equation $(4)$, $c = a - d = 4 - 2 = 2$.

As a result, the values of coefficients would be: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence the balance equation with integer coefficients:

$$\ce{4 XeF4 + 8 H2O -> 2 XeO3 + 2 Xe + 1 O2 + 16 HF}$$