# Reductant ratios

## Problem

A very uniform grade of iron ore is known to be entirely formed from a mixture of different oxides of iron (iron oxide FeO, magnetite Fe$_3$O$_4$ and hematite Fe$_2$O$_3$), although their ratios are not known.

$1$g of this mixture is fully reduced, leaving $0.748$g of pure iron.

Must all three of the oxides indicated be present in the mixture?

Is there a smallest or largest amount of each oxide present?

What possibilities for the percentage composition of the mixture by mass does the experiment indicate?

Can you create a yield of iron which would prove the absence of at least one of the oxides of iron from the ore?

## Getting Started

## Student Solutions

The easiest way to tackle the first part of this problem is to work out the percentage iron (by mass) in each of the iron ores.

FeO$\ \ =\ \frac{56}{56 + 16} = 77.8$ % Fe

Fe$_3$O$_4\ \ =\ \frac{3 \times 56}{(3 \times 56) + (4 \times 16)}= 72.4$ % Fe

Fe$_2$O$_3\ \ =\ \frac{2 \times 56}{(2 \times 56) +(3 \times 16)} =70$ % Fe

The reduction of the mixture reveals that 'on average', the mixture is 74.8 % Fe.

Since the only ore with an iron content higher than this is FeO, it can be seen that the mixture MUST contain some FeO, and additionally either or both of the other two ores.

To get the greatest possible amount of FeO, the mixture would contain only FeO and Fe$_2$O$_3$ since Fe$_2$O$_3$ has the lowest percentage of iron, and will thus need the greatest amount of FeO. We can set up two simultaneous equations to solve:

$n$FeO $+ m$Fe$_2$O$_3 = 1\ $g

$ (n + 2m)$Fe$ = 0.748\ $g

where $n$ and $m$ are in moles

The equations, when putting in the molecular masses, give:

$72n + 160m = 1$

$56(n + 2m) = 0.748$

It is a simple exercise in simultaneous equations to show that:

$n = 8.57 \times 10^{-3}$ moles, and so the maximum mass of FeO is 0.617 g.

To get the smallest allowable mass of FeO, the mixture would contain only FeO and Fe$_3$O$_4$. We can set up two simultaneous equations:

$n$FeO$ + m$Fe$_3$O$_4 = 1\ $g

$(n + 3m)$Fe$ = 0.748\ $g

where $n$ and $m$ are in moles

The equations, when putting the molecular masses, give:

$72n + 232m = 1$

$56(n + 3m) = 0.748$

These can be solved to give:

$n = 6.18 \times 10^{-3}$ moles, and so the minimum mass of FeO is 0.445 g.

Clearly, the minimum amounts of either of the other two ores is zero, since there is only a requirement for either of them to be in the mixture.

Additionally the maximum possible mass of Fe$_2$O$_3$ occurs when there is no Fe$_3$O$_4$ present. For this scenario, the mass of FeO has been calculated as 0.617 g. Therefore this gives the maximum mass of $\mathbf{Fe_2O_3}$ as 0.383 g.

Similarily, the maximum possible mass of Fe$_3$O$_4$ occurs when no Fe$_2$O$_3$ is present. This scenario gives the mass of FeO as 0.445 g, and so the maximum mass of $\mathbf{Fe_3O_4}$ is 0.555 g.

Therefore, the experiment indicates that the mixture can be composed of between 44.5% and 61.7% FeO, with a maximum percentage of 38.3% Fe$_2$O$_3$ and 55.5% Fe$_3$O$_4$. At the upper extreme of FeO composition, the remainder of the mixture is Fe$_2$O$_3$. As the percent FeO decreases, the amount of Fe$_2$O$_3$ also decreases, the the amount of Fe$_3$O$_4$ increases.

There are only two yields of iron which prove the absence of at least one of the oxides. Any yield between 70 and 77.7% exclusive can be made from all three of the oxides, and so none can be excluded. However, a yield of exactly 77.7% can only occur from pure FeO and 70% from pure Fe$_2$O$_3$.

Perhaps what is more interesting is to look at yields which indicate that a certain oxide MUST be present. A yield lower than 72.4% (but higher than 70%) can only be achieved if Fe$_2$O$_3$, whereas a yield greater than 72.4% (but less than 77.7%) can only occur if some FeO is present.