# Question 6fa05

Aug 10, 2016

$\text{0.74 g}$

#### Explanation:

The first thing to do here is to figure out the percent composition of iron in iron(III) oxide, ${\text{Fe"_2"O}}_{3}$, and in iron(II) oxide, $\text{FeO}$, by using the relative atomic masses of the two elements, oxygen and iron.

You will have

$\text{For FeO: " 56/(56 + 16) xx 100 = "77.78% Fe}$

$\text{For Fe"_2"O"_3: " "(56 * 2)/(56 * 2 + 16 * 3) xx 100 = "70.0% Fe}$

Now, let's assume that this mixture contains ${m}_{x}$ grams of iron(II) oxide and ${m}_{y}$ grams of iron(III) oxide. You know that

${m}_{x} + {m}_{y} = 1.0 \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Moreover, you know that the percent composition of iron in this mixture is equal to 72.%, which means that you get $\text{72. g}$ of iron for every $\text{100 g}$ of mixture.

In your case, $\text{1.0 g}$ of mixture will contain

1.0 color(red)(cancel(color(black)("g mixture"))) * "72. g Fe"/(100color(red)(cancel(color(black)("g mixture")))) = "0.72 g Fe"

Use the percent composition of iron in the two oxides to write -- I'm using decimal composition, which is simply percent composition divided by $100$

overbrace(0.7778 * m_x)^(color(blue)("mass of Fe in FeO")) " "+" " overbrace(0.700 * m_y)^(color(purple)("mass of Fe in Fe"_2"O"_3)) = 0.72" " " "color(orange)((2))

Your goal is to find the value of ${m}_{y}$, so use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to write

${m}_{x} = 1.0 - {m}_{y}$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(2\right)}$ to find

$0.7778 \cdot \left(1.0 - {m}_{y}\right) + 0.700 \cdot {m}_{y} = 0.72$

$0.7778 - 0.7778 \cdot {m}_{y} + 0.700 \cdot {m}_{y} = 0.72$

$0.0778 \cdot {m}_{y} = 0.0578 \implies {m}_{y} = \frac{0.0578}{0.0778} = 0.74$

Since ${m}_{y}$ represents the mass of iron(III) oxide present in the mixture, you will have

m_("Fe"_2"O"_3) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.74 g")color(white)(a/a)|)))#

The answer is rounded to two sig figs.