"364.0 g Sb"_2"S"_3" reacts with excess "Fe" to produce "35.0 g Sb". What is the percent yield?

$\text{Sb"_2"S"_3 + "3Fe}$$\rightarrow$$\text{2Sb + 3FeS}$

Mar 14, 2017

The theoretical yield is 13.4%.

Explanation:

$\text{Sb"_2"S"_3" + 3Fe}$$\rightarrow$$\text{2Sb + 3FeS}$

Determine the mole ratios between ${\text{Sb"_2"S}}_{3}$ and $\text{2Sb}$.

$\left(1 \text{mol Sb"_2"S"_3)/(2"mol Sb}\right)$ and $\left(2 \text{mol Sb")/(1"mol Sb"_2"S"_3}\right)$

The mass of $\text{Sb"_2"O"_3}$ needs to be converted to moles. Divide the given mass by its molar mass, $\text{339.7 g/mol}$. https://pubchem.ncbi.nlm.nih.gov/compound/16685273

(364.0color(red)(cancel(color(black)("g"))) "Sb"_2"S"_3)/(339.7color(red)cancel(color(black)("g"))/"mol")="1.0715 mol Sb"_2"S"_3"

Multiply the moles ${\text{Sb"_2"S}}_{3}$ by the mole ratio determined above with $\text{2 mol Sb}$ in the numerator.

1.0715color(red)(cancel(color(black)("mol"))) "Sb"_2"S"_3xx(2 "mol Sb")/(1color(red)(cancel(color(black)("mol") ))"Sb"_2"S"_3)="2.143 mol Sb"

Multiply the moles $\text{Sb}$ by its molar mass to determine the theoretical mass of $\text{Sb}$. Molar mass is the atomic weight on the periodic table in grams/mole, and for Sb it is $\text{121.76 g/mol}$.

2.143 color(red)(cancel(color(black)("mol Sb")))xx(121.76"g Sb")/(color(red)(cancel(color(black)("mol Sb"))))="260.9 g Sb"

Theoretical yield:

"% yield"=("actual yield")/("theoretical yield")xx100

$\text{% yield"=(35.0 color(red)(cancel(color(black)("g"))))/(260.9 color(red)(cancel(color(black)("g"))))xx100="13.4 %}$