Question 0d581

Jun 22, 2017

${\text{FeO " : " Fe"_2"O}}_{3} = 4 : 3$

Explanation:

Start by writing the balanced chemical equations that describe the two reactions

$2 {\text{Fe"_ ((s)) + "O"_ (2(g)) -> 2"FeO}}_{\left(s\right)}$

$4 {\text{Fe"_ ((s)) + 3"O"_ (2(g)) -> 2"Fe"_ 2"O}}_{3 \left(s\right)}$

Now, if you take $x$ to be the number of moles of iron that reacts to form iron(II) oxide and $y$ the number of moles of iron that reacts to form iron(III) oxide, you can say that you have

$x + y = 1 \text{ } \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$

Notice that the first reaction consumes iron and oxygen gas in a $2 : 1$ mole ratio. This implies that the number of moles of oxygen gas consumed to form iron(II) oxide is equal to

x color(red)(cancel(color(black)("moles Fe"))) * "1 mole O"_2/(2color(red)(cancel(color(black)("moles Fe")))) = x/2 ${\text{moles O}}_{2}$

Similarly, the second reaction consumes iron and oxygen gas in a $4 : 3$ mole ratio. This implies that the number of moles of oxygen gas consumed to form iron(III) oxide is equal to

y color(red)(cancel(color(black)("moles Fe"))) * "3 moles O"_2/(4color(red)(cancel(color(black)("moles Fe")))) = (3/4y) ${\text{moles O}}_{2}$

You can thus say that you have

$\frac{x}{2} + \frac{3}{4} y = 0.65 \text{ } \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$

You now have a system of two equations with two unknowns. Use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(1\right)}$ to write

$x = 1 - y$

Plug this into equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(2\right)}$ to get

$\frac{1 - y}{2} + \frac{3}{4} y = 0.65$

$\frac{1}{2} - \frac{2 y}{4} + \frac{3 y}{4} = 0.65$

$\frac{y}{4} = \frac{13}{20} - \frac{10}{20}$

$y = \frac{3}{20} \cdot 4 = \frac{3}{5}$

This means that you have

$x = 1 - \frac{6}{10} = \frac{2}{5}$

You can thus say that the first reaction will produce

2/5 color(red)(cancel(color(black)("moles Fe"))) * "2 moles FeO"/(2color(red)(cancel(color(black)("moles Fe")))) = 2/5 $\text{moles FeO}$

The second reaction will produce

3/5 color(red)(cancel(color(black)("moles Fe"))) * ("2 moles Fe"_2"O"_3)/(4color(red)(cancel(color(black)("moles Fe")))) = 3/10 ${\text{moles Fe"_2"O}}_{3}$

Therefore, the ratio between the iron(II) oxide, or ferrous oxide, and irton(III) oxide, or ferric oxide, will be

"FeO"/("Fe"_2"O"_3) = (2/5 color(red)(cancel(color(black)("moles"))))/(3/10color(red)(cancel(color(black)("moles")))) = 2/5 * 10/3 = 4/3#