# Balancing the oxidative decomposition of FeS2

Doing some practice for the GRE next year.

Please could you look at my working for this and tell me why I'm wrong?

$$\ce{? FeS2 + ? O2 + ? H2O = ? Fe(OH)3 + ? H2SO4}$$

My working so far is:

• There must be the same number of of $\ce{FeS2}$ as $\ce{Fe(OH)3}$

• There must be twice as many $\ce{H2SO4}$ as $\ce{FeS2}$

• If there is one mole of $\ce{FeS2}$ for example, there must be 1 mole of $\ce{Fe(OH)3}$ and two moles of $\ce{H2SO4}$

Tot up the $\ce{H}$ on the right: $1\times 3 + 2\times 2 = 7$. $7$ on the right so $\ce{3.5 H2O}$ on the left. ($7/2 = 3.5$)

$\ce{O}$ on the right = $1\times 3 + 2\times 4 = 11$.

This mean $11$ on the left, too. I already have $\ce{3.5 H2O}$, so we must have $11-3.5 = 7.5$ $\ce{O}$'s from the $\ce{O2}$. So $\ce{3.75 O2}$ because $7.5/2 = 3.75$.

Following the rules I've written down I arrive at:

$$\ce{1.75 FeS2 + 3.75 O2 + 3.5 H2O -> 1.75 Fe(OH)3 + 3.5 H2SO4}$$

I've multiplied these by $4$ to give integers, because that's what the question asks of me, so $7, 15, 14, 7$ and $14$.

Why am I wrong?

• Your equation (if this is it) is not balanced in H and O. Besides, having an acid and a base at the same time among the products does not feel right at all. $$\ce{7 FeS2 + 15 O2 + 14 H2O = 7 Fe(OH)3 + 14 H2SO4}$$ Anyway, welcome to Chem.SE, and pay attention to the formatting. – Ivan Neretin Nov 4 '16 at 16:16
• I've now balanced this equation and the answers are 4, 15, 14 = 4, 8 For anyone who would like to know! – New Zealand's finest Nov 4 '16 at 16:31

I found a solution which works, $$\ce{4FeS_2 + 15O_2 +14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

This is how I found it. Do you know linear algebra ? Because that is very useful. Anyway without to know this is linear algebra you'll be able to understand.

So we are looking about the stoichiometric coefficients $\{\ce{a, b, c, d, e}\}$ of this reaction :

$$\ce{aFeS_2 + bO_2 + cH_2O + d Fe(OH)_3 + eH_2SO_4 \rightarrow 0}$$

Now what you need to do is first to find the number of different elements which are involved in your reaction, here they're four : $\{\ce{\color{\green}{H}, \color{\red}{O}, \color{\purple}{S}, \color{\orange}{Fe}}\}$. I'm not that good in the use of $\LaTeX$ but now the idea is to make column vectors under each of the reagents of your reactions and in these vectors you put in the same order for each the number of each elements there are in and if there is not just put a zero. It will gives you this :

$$a\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{0} \\ \color{\purple}2 \\ \color{\orange}{1} \end{pmatrix}+b\cdot\begin{pmatrix} \color{\green}{0} \\ \color{\red}{2} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+c\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}{1} \\ \color{\purple}{0} \\ \color{\orange}{0} \end{pmatrix}+d\cdot\begin{pmatrix} \color{\green}3 \\ \color{\red}{3} \\ \color{\purple}{0} \\ \color{\orange}1 \end{pmatrix}+e\cdot\begin{pmatrix} \color{\green}2 \\ \color{\red}4 \\ \color{\purple}1 \\ \color{\orange}0 \end{pmatrix}=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}$$

It can be easily transform into a system like this :

$$\begin{cases} 2c+3d+2e=0 \\ 2b+c+3d+4e=0 \\ 2a+e=0 \\ a+d=0 \end{cases}$$

So here we are in a good way because we have more unknowns than equation so if the system converge it has an infinity of solutions. And we can multiply by the number we want each stoichiometric coefficients and still have the same reaction.

So here we will determine four unknowns in terms of an other, I chose $e$. And I got : \begin{cases} a=-e/2 \\ b=-15e/8 \\ c=-7e/4 \\ d=e/2 \\ e=e \end{cases}

I multiply then by $8$ to have only integers which give me with $e=-1$ :

$$\ce{4FeS_2 + 15O_2 + 14H_2O - 4Fe(OH)_3 - 8H_2SO_4 \rightarrow 0}$$

Then $$\ce{4FeS_2 + 15O_2 + 14H_2O = 4Fe(OH)_3 + 8H_2SO_4}$$

I will take pretty much the same approach that 9-BBN does except I will do it a little bit differently. I have never heard of setting the reactants to zero and I'm not very good at making matrices, so I'll just jump right into the systems of equations.

First of all, define your coefficients $a,b,c,d,e$ for each of the chemicals involved, being the reactants that you start with and the products that you get from the reaction.

$$\ce{a FeS2 + b O2 + c H2O -> d Fe(OH)3 + e H2SO4}$$

Now we keep a tally of the number of each atom in molecule. If I put a zero, it means that there are none in that compound and this is for clarity; an equality sign replaces the reaction arrow separating products and reactants.

• Iron (Fe): $\displaystyle 1a + 0b + 0c = 1d + 0e \Longrightarrow a = d$

• Sulfur (S): $\displaystyle 2a + 0b + 0c = 0d + 1e \Longrightarrow 2a = e$

• Oxygen (O): $\displaystyle 0a + 2b + 1c = 3d + 4e \Longrightarrow 2b + c = 3d + 4e$

• Hydrogen (H): $\displaystyle 0a + 0b + 2c = 3d + 2e \Longrightarrow 2c = 3d + 2e$

Now then, we have four equations and 5 variables $a,b,c,d,e$ … so it shouldn't be solvable, except that we only want one solution which is the lowest integer coefficients for each molecule. So we have to make one initial guess for any of the variables. We choose the one that is the simplest and that will be $a$. We set $a = 1$ for initial guess. Note that $a < 0$ & $a = 0$ are forbidden.

In iron (Fe), $a = d$, as $a = 1$, then $1 = d$

In sulfur (S), $2a = e$, as $a = 1$, then $2 = e$

What we know: $a = 1, b =\ ?, c =\ ?, d = 1, e = 2$

With what we know, we can only solve for $c$ in the hydrogen equation next.

$c = [(3d + 2e) / 2]$, as $d = 1, e = 2$, then: $c = 7 / 2$

Solving the last equation for b:

$b = [3d + 4e -c] / 2$, given known variables, then: $b = 15 / 4$

Solutions: $a = 1, b = (15 / 4), c = (7 / 2), d = 1, e = 2$

This will balance the equation according to conservation of mass. However, it makes more sense if you multiply $a$ through $e$ by $4$ so as to get the lowest whole number solutions. This is like scaling a recipe. We are synthesizing the same thing, though.

Therefore, $a = 4, b = 15, c = 14, d = 4, e = 8$

Balanced Chemical Equation:

$$\ce{4 FeS2 + 15 O2 + 14 H2O -> 4 Fe(OH)3 + 8 H2SO4}$$

Now that your equation is balanced I will show you a little something that I've derived. So the formula for the commplete combustion of every hydrocarbon alkane ($\ce{C_nH_{2n+2}}$) such as methane, ethane, propane, butane, pentane, etc … is this:

$$\ce{C_nH_{2n+2} + (3n + 1) / 2 O2 -> n CO2 + (n + 1) H2O}$$

So if we are completely combusting (full airflow of oxygen, no major flickering of the flame producing a mixture of $\ce{CO}$ and $\ce{CO2}$) propane which has the molecular formula $ce{C3H8}$, where $n = 3$ then it's combustion is the following:

$$\ce{C3H8 (g) + 5 O2 (g) -> 3 CO2 (g) + 4 H2O (g)}$$

I know the general complete combustion equation is true for every alkane because all alkanes have the formula $\ce{C_nH_{2n+2}}$, $\ce{O2}$ is always involved in combustion, and $\ce{CO2}$ & $\ce{H2O}$ are always the products of the complete combustion of alkanes. I think this is really cool given that most popular general chemistry equations are combustion equations!

Additional information: I have a lot of ideas running through my head now so once you've determined the balanced chemical equation, you may be able to calculate the spontaneity of the reaction from the change in Gibbs Free Energy of reaction which is useful for knowing if the reaction is a waste of energy, that is that it consumes more energy than produces or not!

You can determine the percent yield after experimentation, how useful your reaction is from the stoichiometric amount of expected product produced. But even with ideally $100~\%$ yield, are all the elements that you find in your desired product going on to form your product or are they being wasted producing something else?

The 2nd Principle of Green Chemistry: Atom Economy, states that

Synthetic methods should be designed to maximize incorporation of all materials used in the process into the final product. (The American Chemical Society)

So what's the point of your reaction? To make $\ce{Fe(OH)3}$? What percent of the atoms $\ce{Fe, O}$, and $\ce{H}$ are being incorporated as $\ce{Fe(OH)3}$ and how much is going off to form sulfuric acid instead? We can calculate that as the percent atom economy.

$$\%\ \text{Atom Economy} = \frac{\text{mass of atoms in desired product}}{\text{mass of atoms in all reactants}} \times 100$$

where units are in grams per mole, but they are eliminated in the ratio.

\begin{align}\%\ \text{Atom Economy} &= \frac{4 \times 106.866}{4 \times 119.965 + 15 \times 31.998 + 14 \times 18.015} \times 100\\ &= \frac{427.464}{1212.04} \times 100\\ &= 35.27~\%\end{align}

So if there were multiple ways to make $\ce{Fe(OH)3}$ $\%\ \text{Atom economy}$ may be a useful factor to consider when conducting a synthesis. I think this is more important, though, in choosing a synthetic reaction where toxic byproducts are involved and you want to minimize the toxic byproducts produced for the health of the earth and of the customer or patient and also to reduce on having to spend money to deactivate and/or filter out toxic byproducts in an industrial synthesis. Therefore, you want the synthesis with the highest $\%\ \text{Atom Economy}$.

• I think that this chemical equation solver can solve your difficult equation in flash. It probably uses either matrices or systems of equations. You can use this if you like: webqc.org/balance.php ... anyway, it's for speed of calculation. It doesn't not explain the methodology like I did, though. – xyz123 Nov 5 '16 at 7:35
• I don't believe that I said that your mathematics is unnecessary, just that I would do it a little bit differently. – xyz123 Nov 5 '16 at 16:50
• You cannot comment everywhere because that is a priviledge associated with 50 reputation. Rather, since this was a simple mathematical error, you could have proposed an edit instead. I will now read on to review the remainder of your post to determine whether it answers the question or not (only up to the third paragraph). – Jan Nov 5 '16 at 18:27
• Okay, more than $50~\%$ of your post ramble on about things that are not relevant to the question. Also, I learnt a different definition of atom economy, but that is not important here. – Jan Nov 5 '16 at 18:47
• What is the definition of atom economy that you learned? – xyz123 Nov 5 '16 at 19:15

This is a redox equation. You can approach it in a standard redox manner, however, you will need to use a small, helpful trick.

If we set up redox pairs, we will notice that there are two compounds that get oxidised but only one that gets reduced:

\begin{align}\ce{Fe^2+} &/ \ce{Fe^3+}\tag{Ox1}\\ \ce{(S^{-I})2^2-} &/ \ce{(S^{+VI})O4^2-}\tag{Ox2}\\ \ce{(O^{\pm 0})2} &/ \ce{(O^{-II})H-}\tag{Red}\end{align}

However, we can alleviate this problem by realising that $\ce{FeS2}$ is likely a mineral and thus we need to oxidise one iron per $\ce{S2^2-}$ unit. Also, rather than considering water it seems the question wants you to generate hydroxide, hence why $\ce{OH-}$ is my reduced partner of elemental oxygen. That lets us set up the half equations. The reduction is first because it is a lot simpler.

\begin{align}\ce{O2 + 4e- + 2 H+ &-> 2 OH-}\tag{Red}\\ \strut\\ \ce{FeS2 \phantom{\ce{+ 8 H2O}} &-> Fe^3+ + 2 SO4^2- + 15 e-}\tag{Ox1}\\ \ce{FeS2 \phantom{\ce{+ 8 H2O}} &-> Fe^3+ + 2 SO4^2- + 15 e- + 16 H+}\tag{Ox2}\\ \ce{FeS2 + 8 H2O &-> Fe^3+ + 2 SO4^2- + 15 e- + 16 H+}\tag{Ox3}\end{align}

The smallest common multiple of $4$ and $15$ happens to be $60$, thus $\text{(Ox3)}$ must be multiplied with $4$ and $\text{(Red)}$ must be multiplied with $15$ leading us to $\text{(RO1)}$ and the following:

\begin{align}\ce{4 FeS2 + 32 H2O + 15 O2 + 30 H+ &->\\ 4 Fe^3+ &+ 8 SO4^2- + 64 H+ + 30 OH-}\tag{RO1}\\ \ce{4 FeS2 + 32 H2O + 15 O2 &->\\ 4 Fe^3+ &+ 8 SO4^2- + 34 H+ + 30 OH-}\tag{RO2}\\ \ce{4 FeS2 + 32 H2O + 15 O2 &->\\ 4 Fe^3+ &+ 8 SO4^2- + 30 H2O + 4 H+}\tag{RO3}\\ \ce{4 FeS2 + 2 H2O + 15 O2 &->\\ &4 Fe^3+ + 8 SO4^2- + 4 H+}\tag{RO4}\end{align}

Everything henceforth is just adding further water molecules or regrouping at your desire. For example, to me, the equation suggests combining the free protons with sulfate ions and even the iron cations in the following way $\text{(RO5)}$:

$$\ce{4 FeS2 + 2 H2O + 15O2 -> 4 Fe(SO4)(HSO4)}\tag{RO5}$$

On the other hand, if you really had to include the generation of sulfuric acid and iron hydroxyide, note that you are lacking twelve hydroxide ions ($4 \times 3$) and 12 protons ($8 \times 2 - 4$) — meaning that twelve water molecules should be added to the left-hand side.

$$\ce{4 FeS2 + 14 H2O + 15 O2 -> 4 Fe(OH)3 + 8 H2SO4}\tag{RO6}$$

Although I must agree with Ivan’s comment: the co-presence of $\ce{H2SO4}$ and $\ce{Fe(OH)3}$ is extremely unlikely; the following will happen:

$$\ce{2Fe(OH)3 (s) + 3H2SO4 (aq) -> 2 Fe^3+ (aq) + 6 H2O (l) + 3 SO4^2- (aq)}$$

I used to be pretty good at balancing chemical equations, but now days I'm pretty rusty. Never-the-less, as far as I can see the equation you've been given must be wrong (or you mis-wrote it above). FeS2 would be assumed by a beginning student to have a ferric (+IV) ion (formally, at least) combined with two sulfide ions S=. So your reaction reduces the Iron (to Fe(+III)) and oxidizes the sulfurs (to S(+VI)). The oxidation of the sulfurs in this scenario gives you all the electrons you need to reduce the oxygen to the oxide (O=) for the formation of the sulfate anion. In other words you'd be left with an "extra" electron (from the iron) - which as long as this isn't a half reaction isn't going to happen (at least as far as elementary chemical homework is concerned). A more sophisticated student might learn that Sulfur commonly forms more exotic ions, such as the disulfide anion S-S= and that the two common forms of FeS2 both have Iron in the +2 not the +4 valence state. This means iron goes from +2 to +3 in your given reaction. Which is pretty much the same problem (in reverse, actually): the sulfur oxidation gives you 14 electrons, the Fe oxidation 1 more and you need 16 to reduce the oxygen a discrepancy of one electron. There's no place for that electron to go! You're stuck, again.