# In this reaction: 4HCl (g) + O_2 (g) -> 2H_2O (l) + 2Cl_2(g). When 63.1 g of HCI react with 17.2 g of O_2, 49.3 g of Cl_2 are collected. How do you determine the limiting reactant, theoretical yield of Cl_2 and percent yield for the reaction?

Jul 17, 2016

WARNING! Long answer. The limiting reactant is $\text{HCl}$. The theoretical yield of ${\text{Cl}}_{2}$ is 61.4 g. The percentage yield is 80.3 %

#### Explanation:

We must first identify the limiting reactant, and then we calculate the theoretical yield and percent yields.

$\textcolor{w h i t e}{m m m m m m m m} {\text{4HCl" + "O"_2 → "2H"_2"O" + "2Cl}}_{2}$
$\text{MM/g·mol"^"-1} : \textcolor{w h i t e}{m l} 36.46 \textcolor{w h i t e}{m} 32.00 \textcolor{w h i t e}{m m m m m l} 70.91$

(a) Identify the limiting reactant

We calculate the amount of chlorine that can form from each reactant.

Calculate the moles of $\text{HCl}$

$\text{Moles of HCl" = 63.1 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "1.731 mol HCl}$

Calculate moles of ${\text{Cl}}_{2}$ formed from the $\text{HCl}$

${\text{Moles of Cl"_2 = 1.731color(red)(cancel(color(black)("mol HCl"))) × "2 mol Cl"_2/(4 color(red)(cancel(color(black)("mol HCl")))) = "0.8653 mol Cl}}_{2}$

Calculate the moles of ${\text{O}}_{2}$

${\text{Moles of O"_2 = 17.2 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.5375 mol O}}_{2}$

Calculate the moles of ${\text{Cl}}_{2}$ formed from the ${\text{O}}_{2}$

${\text{Moles of Cl"_2 = 0.5375 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol Cl"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "1.075 mol Cl}}_{2}$

The limiting reactant is $\text{HCl}$, because it gives fewer moles of ${\text{Cl}}_{2}$.

(b) Calculate the theoretical yield of ${\text{Cl}}_{2}$.

${\text{Theoretical yield" = 0.8653 color(red)(cancel(color(black)("mol Cl"_2))) × "70.91 g Cl"_2/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "61.4 g Cl}}_{2}$

The theoretical yield of ${\text{Cl}}_{2}$ is 61.4 g.

(c) Calculate the percentage yield of ${\text{Cl}}_{2}$

The formula for percentage yield is

color(blue)(|bar(ul(color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "

"% yield" = (49.3 color(red)(cancel(color(black)("g"))))/(61.4 color(red)(cancel(color(black)("g")))) × 100 % = 80.3 %