Stoichiometry

Applications

It is great that we are able to balance chemical equations, but how do we actually use this information? Stoichiometric coefficients have a wide array of uses however the most commonly tested questions focus on limiting reagents, determining the reaction yield, and how much of a reactant is leftover.

Limiting Reagent

The limiting reagent is the reactant that “holds the reaction back”. Basically, it places a limit on how much product can be generated because this reactant runs out all the other reactants have been used up. Let’s look at the combustion reaction we just balanced to see how this works.

Let’s say that we have 4 mols of O₂ and 1 mol of CH₄ that are going to react. By using our mole to mole ratios and dimensional analysis we can determine how much CO2 and water is produced by the oxygen.

[latexpage]
To do this we will use the general dimensional analysis setup

\[ \#\; mols\; Reactant \times \frac{Product\;Coeffcient}{Reactant\;Coeffcient} = Product\; Produced\; Mols
\]
\[ 4\; mols\; O_2 \times \frac{1\;mol\;CO_2}{2\;mols\;O_2} = 2\; mols\; CO_2
\]

\[ 4\; mols\; O_2 \times \frac{2\;mols\;H_2O}{2\;mols\;O_2} = 4\; mols\; O_2
\]

So if only 4 moles of oxygen were present and an unlimited amount of methane (CH4) was present the reaction would produce 2 mols of CO₂ and 4 moles of H₂O.

However, there is only one mol of methane present so we need to see if this is going to “hold the reaction back”. Again we will use our mol to mol ratios to determine how many mols of CO₂ and water are produced when one mol of CH₄ reacts with unlimited oxygen.

[latexpage]
\[ 1\; mols\; CH_4 \times \frac{1\;mol\;CO_2}{1\;mols\;CH_4} = 1\; mols\; CO_2
\]

\[ 4\; mols\; CH_4 \times \frac{2\;mols\;H_2O}{2\;mols\;CH_4} = 2\; mols\; O_2
\]

Therefore, when we have 1 mol of CH₄ and 4 mols of O₂ the methane will limit how much CO₂ and H₂O is produced. In essence, all of the methane is used up before the oxygen gas is. Therefore methane is the limiting reagent.

What if we weren’t given the number of moles, but instead were given the values of each reactant in grams. We will still go through the exact same steps as above except we would need to convert from gram to mols first using the molar mass of the reactant.

Limiting Reagent Summary

1. Balance the reaction
2. Convert all reactant values to mols using the molar masses of each reactant
3. Using dimensional analysis convert all reactants into one of the products
4. The reactant that produces the smallest number of mols of product is the limiting reagent

Leftovers

As we have seen above some reactants aren’t completely consumed during the reaction. This will always be the non-limiting reactant unless the perfect amount of each reactant is present. In this last scenario, only one methane is present so only two oxygen molecules will be able to react. This leaves two unreacted oxygen gas molecules behind. In many reactions, there are leftovers although the ratios don’t usually work out quite as nicely.

To calculate this you can convert the amount of product that is actually produced back into the non-limiting reactant again by using dimensional analysis. This value will represent the amount of reactant that was used up in the process of the reaction. To find the amount of leftover reactant we can then subtract this value from what we started with.

[latexpage]
\[
Amount\; Leftover = Starting\; Amount – Amount\; Used
\]

Yield

Lastly, we can also calculate a reaction’s predicted yield or theoretical yield. Theoretical yield refers to how much product would be produced if the reaction went 100% to products. We have already learned how to calculate this value although we might not realize it. When we calculated the limiting reagent we converted our reactants into products. The value of that product is the same as our theoretical yield. Again if we needed that value in grams and not mols we could use the molar mass to convert our units to grams.

In real life, reactions don’t react perfectly and not all of the reactants will end up as products. The value of the products produced when the actual experiment is carried out is called the actual yield and by comparing the two values to one another we can determine how efficient our reaction was.

For example, if our reaction only produced 13g of CO₂ then we know our yield isn’t 100% since we predicted 22g of CO₂ being produced. If we want to quantify the efficiency of this reaction we can by calculating the percent yield.

[latexpage]
\[
Percent\; Yield = (\frac{Actual\;Yield}{Theoretical\;Yield})\times 100\%
\]