# Given the following, what is the limiting reactant? What is the theoretical yield of Cl_2? If the yield of the reaction is 75.5%, what is the actual yield of chlorine?

## Chlorine gas can be prepared in the laboratory by the following reaction: $4 H C l \left(a q\right) + M n {O}_{2} \left(s\right) \to M n C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right) + C {l}_{2} \left(g\right)$. You add 40.5 g of $M n {O}_{2}$ to a solution containing 41.7 g $H C l$.

Nov 20, 2016

The actual yield is 3.83 g of $\text{Cl"_2}$.

#### Explanation:

This is a limiting reactant problem.

We know that we will need a balanced equation with masses, molar masses, and moles of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and everything else below the formulas.

${M}_{r} : \textcolor{w h i t e}{m m m m} 36.46 \textcolor{w h i t e}{m l} 86.94 \textcolor{w h i t e}{m m m m m m m m m m l l} 70.91$
$\textcolor{w h i t e}{m m m m m m} {\text{4HCl" + "MnO"_2 → "MnCl"_2 + "2H"_2"O" + "Cl}}_{2}$
$\text{Mass/g:} \textcolor{w h i t e}{m m l} 41.7 \textcolor{w h i t e}{m m l} 40.5$
$\text{Amt/mol:} \textcolor{w h i t e}{m l} 1.144 \textcolor{w h i t e}{m} 0.4658$
$\text{Divide by:} \textcolor{w h i t e}{m m} 4 \textcolor{w h i t e}{m m m m} 1$
$\text{Moles rxn:} \textcolor{w h i t e}{l l} 0.2859 \textcolor{w h i t e}{m} 0.4658$

2. Identify the limiting reactant

An easy way to identify the limiting reactant is to calculate the "moles of reaction" each will give:

You divide the moles of each reactant by its corresponding coefficient in the balanced equation.

I did that for you in the table above.

$\text{HCl}$ is the limiting reactant because it gives the fewest moles of reaction.

3. Calculate the theoretical moles of ${\text{Cl}}_{2}$

${\text{Theoretical yield" = 0.2859 color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Cl"_2)/(4 color(red)(cancel(color(black)("mol HCl")))) = "0.071 48 mol Cl}}_{2}$

4. Calculate the theoretical yield of ${\text{Cl}}_{2}$

${\text{Theoretical yield" = "0.071 48" color(red)(cancel(color(black)("mol Cl"_2))) × ("70.91 g Cl"_2)/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "5.069 g Cl}}_{2}$

5. Calculate the actual yield

$\text{Actual yield" = 5.069 color(red)(cancel(color(black)("g theoretical"))) × ("75.5 g actual")/(100 color(red)(cancel(color(black)("g theoretical")))) × 100 % = "3.83 g}$

The actual yield of ${\text{Cl}}_{2}$ is 3.83 g.