Question #6fa05

1 Answer
Aug 10, 2016

Answer:

#"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.