We start with the balanced equation.
#"M_r:color(white)(mmmm) 231.53color(white)(I'll) 28.01#
#color(white)(mmml) "4C" + "Fe"_3"O"_4 → 4"CO" + "3Fe"#
For convenience, I have written the molar masses above the formulas.
(a) Calculate the moles of #"Fe"_3"O"_4#
#"Moles of Fe"_3"O"_4 = 25 color(red)(cancel(color(black)("g Fe"_3"O"_4))) × ("1 mol Fe"_3"O"_4)/(231.53 color(red)(cancel(color(black)("g Fe"_3"O"_4)))) = "0.108 mol Fe"_3"O"_4#
(b) Calculate moles of #"CO"# formed from the #"Fe"_3"O"_4#
#"0.108"color(red)(cancel(color(black)("mol Fe"_3"O"_4))) × ("4 mol CO")/(1 color(red)(cancel(color(black)("mol Fe"_3"O"_4)))) = "0.432 mol CO"#
(c) Calculate the theoretical yield of #"CO"#.
#"Theoretical yield" = 0.432 color(red)(cancel(color(black)("mol CO"))) × ("28.01 g CO")/(1 color(red)(cancel(color(black)("mol CO")))) = "12.10 g CO""#
The theoretical yield of #"CO"# is 12.10 g.
(d) Calculate the percent yield of #"CO"#.
The formula for percentage yield is
#color(blue)(bar(ul(|color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "#
#"Percent yield" = (9.8 color(red)(cancel(color(black)("g"))))/(12.10 color(red)(cancel(color(black)("g")))) × 100 % = 81 %#
The percent yield is 81 %.
