Limiting Reagents, Theoretical Yield, and Actual Yield

A balanced chemical reaction allows prediction of product yield based on reactant amounts.
Limiting reagents become important when initial amounts of two or more reactants are known.
The limiting reagent is the reactant that is in short supply and determines the maximum amount of product that can be formed.
Once the limiting reagent is used up, the reaction stops.

Illustrates the concept of limiting reagents.
A turkey sandwich requires one slice of turkey and two slices of bread.
If you have five slices of turkey and eight slices of bread, you can only make four sandwiches because you run out of bread first.
Bread is the limiting reagent, and turkey is the excess reagent.

Problem: 50 grams of nitrogen react with 7.5 grams of hydrogen to form ammonia (NH3).

Determine the limiting reagent by comparing the amount of product (ammonia) that can be made from each reactant.
First, write and balance the chemical reaction: $N2 + 3H2 \rightarrow 2NH_3$
Convert the mass of each reactant to moles:
- Nitrogen: 50 grams / 28 grams per mole = 1.79 moles of $N_2$
- Hydrogen: 7.5 grams / 2 grams per mole = 3.75 moles of $H_2$
Calculate the amount of ammonia that can be produced from each reactant:
- From Nitrogen: $1.79 \, moles \, N2 \times \frac{2 \, moles \, NH3}{1 \, mole \, N2} = 3.58 \, moles \, NH3$
- From Hydrogen: $3.75 \, moles \, H2 \times \frac{2 \, moles \, NH3}{3 \, moles \, H2} = 2.5 \, moles \, NH3$
The reactant that produces the least amount of product is the limiting reagent. In this case, hydrogen is the limiting reagent.

The largest amount of ammonia that can be formed is determined by the limiting reagent.
From Part 1, we know that hydrogen is the limiting reagent and can produce 2.5 moles of ammonia.
Convert moles of ammonia to grams: 2.5 moles $\times$ 17 grams per mole = 42.5 grams of $NH_3$ (approximately).

The excess reactant is nitrogen.
Determine the amount of nitrogen used in the reaction by using the amount of ammonia formed.
Since 2.5 moles of ammonia are formed, we can calculate the amount of nitrogen used:
- $2.5 \, moles \, NH3 \times \frac{1 \, mole \, N2}{2 \, moles \, NH3} = 1.25 \, moles \, N2$
Subtract the amount of nitrogen used from the initial amount of nitrogen:
- 1. 79 moles (initial) - 1.25 moles (used) = 0.54 moles of $N_2$ left over.
Convert the moles of nitrogen left over to grams:
- 0. 54 moles $\times$ 28 grams per mole = 15.12 grams of $N_2$ left over (approximately).

Theoretical yield is the maximum amount of product that can be made, as predicted by stoichiometry.
Actual yield is the amount of product that is actually obtained from a reaction.
In reality, the actual yield is almost always less than the theoretical yield due to various factors like byproducts, side reactions, incomplete reactions or reverse reactions.
Scientists often compete to achieve actual yields closest to the theoretical yield.
Percent yield is the ratio of actual yield to theoretical yield, expressed as a percentage.

$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$

The combustion of 3 moles of $C2H4$ yields 206 grams of $CO_2$ . What is the percent yield?
Write and balance the combustion reaction: $C2H4 + 3O2 \rightarrow 2CO2 + 2H_2O$
The actual yield of $CO_2$ is given as 206 grams.
Calculate the theoretical yield of $CO_2$ :
- From the balanced equation, 1 mole of $C2H4$ produces 2 moles of $CO_2$ .
- Therefore, 3 moles of $C2H4$ will produce 6 moles of $CO_2$ .
Convert moles of $CO_2$ to grams:
- 6 moles $\times$ 44 grams per mole = 264 grams of $CO_2$
Calculate the percent yield:
- $\text{Percent Yield} = \frac{206 \, grams}{264 \, grams} \times 100\% = 78.03\%$