If 20 g of iron reacted with oxygen to form 20 g of iron(III) oxide, what is the percent of iron reacted?

1 Answer
Sep 3, 2017

Answer:

The percent of iron reacted is 70 %.

Explanation:

Step 1. Write the chemical equation

#M_text(r):color(white)(mm)55.84color(white)(mmmmll)159.69#
#color(white)(mmmmll)"2Fe" + "3O"_2 → "Fe"_2"O"_3#
#"Mass/g:"color(white)(m)20color(white)(mmmmmmll)20#

Step 2. Calculate the moles of #"Fe"_2"O"_3#

#"Moles of Fe"_2"O"_3 = 20 color(red)(cancel(color(black)("g Fe"_2"O"_3))) × ("1 mol Fe"_2"O"_3)/(159.69 color(red)(cancel(color(black)("g Fe"_2"O"_3)))) = "0.125 mol Fe"_2"O"_3#

Step 3. Calculate the moles of #"Fe"#

#"Moles of Fe" = 0.125 color(red)(cancel(color(black)("mol Fe"_2"O"_3))) × "2 mol Fe"/(1 color(red)(cancel(color(black)("mol Fe"_2"O"_3)))) = "0.250 mol Fe"#

Step 4. Calculate the mass of #"Fe"#

#"Mass of Fe" = 0.250 color(red)(cancel(color(black)("mol Fe"))) × "55.84 g Fe"/(1 color(red)(cancel(color(black)("mol Fe")))) = "14.0 g Fe"#

Step 5. Calculate the percent of #"Fe"# reacted

#"% Fe reacted" = "mass of Fe reacted"/"original mass of Fe" × 100 % = (14.0 color(red)(cancel(color(black)("g"))))/(20 color(red)(cancel(color(black)("g")))) × 100 % = 70 %#