Question #6fa05
1 Answer
Explanation:
The first thing to do here is to figure out the percent composition of iron in iron(III) oxide,
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
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
#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.