Question

Question #6fa05

Answer 1

`"0.74 g"`

Explanation:

The first thing to do here is to figure out the percent composition of iron in iron(III) oxide, `"Fe"_2"O"_3`, and in iron(II) oxide, `"FeO"`, by using the 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` grams of iron(II) oxide and `m_y` grams of iron(III) oxide. You know that

`m_x + m_ y= 1.0" " " "color(orange)((1))`

Moreover, you know that the percent composition of iron in this mixture is equal to `72.%`, which means that you get `"72. g"` of iron for every `"100 g"` of mixture.

In your case, `"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 `color(orange)((1))` to write

`m_x = 1.0 - m_y`

Plug this into equation `color(orange)((2))` to find

`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 `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.