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?

Md Ariful Islam

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Md Ariful Islam

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

We start with the balanced equation.

$color(white)(mmmmmmmm)"4HCl" + "O"_2 → "2H"_2"O" + "2Cl"_2$
$"MM/g·mol"^"-1":color(white)(ml) 36.46color(white)(m) 32.00color(white)(mmmmml) 70.91$

(a) Identify the limiting reactant

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

Calculate the moles of $"HCl"$

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

Calculate moles of $"Cl"_2$ formed from the $"HCl"$

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

Calculate the moles of $"O"_2$

$"Moles of O"_2 = 17.2 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.5375 mol O"_2$

Calculate the moles of $"Cl"_2$ formed from the $"O"_2$

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

The limiting reactant is $"HCl"$, because it gives fewer moles of $"Cl"_2$.

(b) Calculate the theoretical yield of $"Cl"_2$.

$"Theoretical yield" = 0.8653 color(red)(cancel(color(black)("mol Cl"_2))) × "70.91 g Cl"_2/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "61.4 g Cl"_2$

The theoretical yield of $"Cl"_2$ is 61.4 g.

(c) Calculate the percentage yield of $"Cl"_2$

The formula for percentage yield is

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

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

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