Question 6009b

Jul 2, 2017

The percent yield is 81 %.

Explanation:

"M_r:color(white)(mmmm) 231.53color(white)(I'll) 28.01
$\textcolor{w h i t e}{m m m l} \text{4C" + "Fe"_3"O"_4 → 4"CO" + "3Fe}$

For convenience, I have written the molar masses above the formulas.

(a) Calculate the moles of ${\text{Fe"_3"O}}_{4}$

${\text{Moles of Fe"_3"O"_4 = 25 color(red)(cancel(color(black)("g Fe"_3"O"_4))) × ("1 mol Fe"_3"O"_4)/(231.53 color(red)(cancel(color(black)("g Fe"_3"O"_4)))) = "0.108 mol Fe"_3"O}}_{4}$

(b) Calculate moles of $\text{CO}$ formed from the ${\text{Fe"_3"O}}_{4}$

$\text{0.108"color(red)(cancel(color(black)("mol Fe"_3"O"_4))) × ("4 mol CO")/(1 color(red)(cancel(color(black)("mol Fe"_3"O"_4)))) = "0.432 mol CO}$

(c) Calculate the theoretical yield of $\text{CO}$.

$\text{Theoretical yield" = 0.432 color(red)(cancel(color(black)("mol CO"))) × ("28.01 g CO")/(1 color(red)(cancel(color(black)("mol CO")))) = "12.10 g CO}$

The theoretical yield of $\text{CO}$ is 12.10 g.

(d) Calculate the percent yield of $\text{CO}$.

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 %.