# Question #6fa05

##### 1 Answer

#### Explanation:

The first thing to do here is to figure out the **percent composition** of iron in iron(III) oxide, *relative atomic masses* of the two elements, oxygen and iron.

You will have

#"For FeO: " 56/(56 + 16) xx 100 = "77.78% Fe"#

#"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 + m_ y= 1.0" " " "color(orange)((1))#

Moreover, you know that the **percent composition** of iron in this mixture is equal to **for every**

In your case,

#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

#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_x = 1.0 - m_y#

Plug this into equation

#0.7778 * (1.0 - m_y) + 0.700 * m_y = 0.72#

#0.7778 - 0.7778 * m_y + 0.700 * m_y = 0.72#

#0.0778 * m_y = 0.0578 implies m_y = 0.0578/0.0778 = 0.74#

Since *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**.