|The image above depicts this reaction:|
3H2 + N2 2NH3
Your teacher needs to make cookies for a bake sale and so has decided to quadruple the recipe. Supplies are low, however, and even though there is plenty of flour and white sugar, etc., there is only enough brown sugar to triple the recipe. So the recipe is tripled, not quadrupled. The amount of cookies made depends on the ingredient that is in shortest supply. When the brown sugar is used up, no more cookies can be made.
Here is an example of the same idea in a chemical reaction. Nitrogen can be reacted with hydrogen to make ammonia:
At the molecular level this reaction can be illustrated as seen at right. If you double the number of nitrogen molecules, then you have to double the number of hydrogen molecules in order to react all of the nitrogen. This is because of the stoichiometric calculation:
3 H2 2 N2 × -------- = 6 H2 1 N2
|The image above depicts this reaction:|
6H2 + 2N2 4NH3
An illustration of this situation is shown at right. Notice that when nitrogen molecules are doubled, so is the number of ammonia molecules. But what if when the number of nitrogen molecules is doubled the number of hydrogen molecules is not doubled? Say there are 2 N2 molecules but only 3 H2 molecules. What happens then? The answer is that there is too much nitrogen for the amount of hydrogen. One molecule of N2 will react but one will be left over because there are no more H2 molecules to react with it. This situation, and its results, are shown below.
|This illustration shows what happens when 3 molecules of hydrogen are reacted with 2 molecules of nitrogen. Two molecules of ammonia form and one molecule of nitrogen is left over.|
The scenarios described above introduce and illustrate the concept of the limiting reactant. A limiting reactant is the chemical for which the supply is less than the molar ratio required in the balanced chemical equation. Consider the combustion of propane:
If a problem states that one reactant is ‘unlimited’ it means that the other reactant is the limiting reactant. Take the case of unlimited oxygen: how much carbon dioxide is produced when all of the propane in a typical propane tank is burned? Typically, a propane tank holds about 4.6 gallons of liquefied propane. Since it has a density of 4.11 pounds per gallon (0.493 g/cm3) that means that the tank holds 8.5 kg (8,500 g). The molar mass of propane is 44.096 g/mol so 8.5 × 103 g is 190 moles of propane. Clearly, the propane is the limiting reactant in this case since oxygen from the air is basically unlimited. So how many moles of carbon dioxide would be produced by burning all of the propane in the tank?
3 mol CO2 190 mol C3H8 × ---------- = 570 mol CO2 1 mol C3H8
Similarly, the amount of water that is produced depends only on the amount of propane burnt. So far so good. But what if the problem does not say which reactant is in excess? How can you decide which one it is?
The reaction above shows what happens when copper metal is dissolved in nitric acid. How many moles of nitrogen monoxide gas (NO) are made when 2.00 mol copper (Cu) are placed in a solution containing 5.00 mol nitric acid (HNO3)? In order to answer the question, we have to know which chemical is the limiting reactant. In the balanced chemical equation for the reaction there is an 8:3 ratio for nitric acid to copper metal. The given amounts in the question have a 5:2 ratio. See below:
|Ratio in Equation||Actual Ratio of Materials|
8 mol HNO3 ------------ 3 mol Cu
5.00 mol HNO3 ------------ 2.00 mol Cu
As you can see, the actual ratio of the materials is less than the molar ratio from the chemical equation. This means that the amount of nitric acid is too small to react with all of the copper. So nitric acid is the limiting reactant and the amount of nitrogen monoxide gas produced can be calculated as follows:
2 mol NO 5.00 mol HNO3 × ----------- = 1.25 mol NO 8 mol HNO3
How much of the excess reactant is left over? Copper was in excess in this case and the amount that reacted with the nitric acid is:
3 mol Cu 5.00 mol HNO3 × ----------- = 1.88 mol Cu 8 mol HNO3
So the amount of copper left over is 2.00 – 1.88 = 0.12 mol Cu.
It is often the case that limiting reactant problems will give only the masses of the reactants, not the number of moles. This is simple enough to handle: just convert all quantities to moles before trying to find which one is the limiting reactant. Once you know how many moles of each reactant you have you can compute the molar ratio between them and compare it to the ratio from the balanced equation. Here is another example to demonstrate the idea:
Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide, water, and calcium chloride. If you have 2.50 g of calcium carbonate and a solution containing 0.100 moles of hydrochloric acid then how many grams of carbon dioxide would be produced?
The first thing to do is to write a balanced equation:
The number of moles of calcium carbonate can be calculated as follows:
1 mol 2.50 g CaCO3 × ---------- = 0.0250 mol CaCO3 100.09 g
Next, determine which reactant is the limiting reactant:
|Ratio in Equation||Actual Ratio of Materials|
2 mol HCl ------------ 1 mol CaCO3
0.100 mol HCl ------------ 0.0250 mol CaCO3
The actual ratio of materials is greater than the molar ratio in the chemical equation. This means that the amount of HCl is larger than necessary to react with all of the CaCO3. This means that CaCO3 is the limiting reactant and the number of moles of CaCO3 should be used to calculate the amount of CO2 that is generated.
1 mol CO2 44.01 g 0.0250 mol CaCO3 x ------------ x --------- = 1.10 g CO2 1 mol CaCO3 1 mol CO2
An alternative way to do the problem is to figure out the limiting reactant after using both reactants to predict the amount of product. Whichever reactant produces less product was the limiting reactant. For example, the mass of CO2 produced based on the number of moles of CaCO3 is 1.10 g (as seen above). The mass of CO2 that would form if all the hydrochloric acid reacted would be:
1 mol CO2 44.01 g 0.100 mol HCl x ------------ x ---------- = 2.20 g CO2 2 mol HCl 1 mol CO2
The idea of the limiting reactant is that it limits the amount of product you can make since you run out of one of the reactants before you run out of the other(s). In the example above, the calcium carbonate gets used up and there is some hydrochloric acid left over. You can tell this is the case in the problem by using each reactant to predict the amount of the product. Whichever reactant produces the smaller amount of product is the limiting reactant.
Up to this point in the discussion of stoichiometry the assumption has been that all reactions go until the limiting reactant is used up. A chemist describes this by saying that the reaction goes to completion. When a reaction goes to completion it uses up all of the limiting reactant and makes only the products shown in the reaction of interest. For some reactions this is absolutely what is observed. When this happens the reaction is said to produce the theoretical yield. The theoretical yield is the maximum amount of product that can be made from a given amount of reactant chemicals. Up to now all the stoichiometric calculations you have done have resulted in an amount of product properly understood as the theoretical yield.
For the majority of chemical reactions it is not the case that the maximum possible amount of product is made. First of all, the written reaction under consideration may not be the only reaction that occurs. When this is the case the reactants participate in side reactions. Side reactions are chemical reactions which may or may not be known but which use up the same reactants as in the chemical equation that you are concerned with. They produce products other than the ones of interest. The consequence of this is that the amount of product actually made is less than the theoretical yield.
Secondly, chemical reactions may be subject to the condition of equilibrium. It may be that the most stable mixture of reactants and products is not one in which the reactants are used up completely to make products. Many chemical reactions are reversible: products can react to make the reactants again. When the equilibrium point is reached the reaction continues at the molecular level but appears to stop based on simple observation. Reactant molecules continue to become product molecules but they do so at the same rate that product molecules react to become reactant molecules. Equilibrium is an advanced topic in chemistry and is beyond the scope of this text. Nevertheless, it is an important contributor to the situation in which the theoretical yield of products is not achieved in experiment.
Lastly, if a chemical reaction is carried out under conditions that are not ideal then the full theoretical yield may not be achieved. For example, some reactions are more efficient at turning reactants into products at high temperatures and other reactions are more efficient at low temperatures.
When the amount of product is less than the theoretical yield it is possible to calculate a percent yield for the reaction. The percent yield is a measure of the efficiency of the reaction in producing the intended product. It is calculated by dividing the actual amount of product by the theoretical yield of product and multiplying by 100%. Here is an example:
In Example 1 the amount of nitrogen monoxide formed was 1.25 mol. This was the theoretical yield for that problem because it was calculated using the limiting reactant. If the experiment were performed in a lab and the actual number of moles of NO was found to be 1.00 mol then what was the percent yield?
1.00 mol ---------- × 100% = 80% 1.25 mol
The percent yield in this case was 80%. This means that only 80% of the theoretical yield was actually collected when the experiment was carried out. Either a side reaction used up some of the nitric acid or the equilibrium amount of NO is less than if the reaction were to go to completion.
Percent yields can also be calculated based on masses.
In Example 2 the theoretical yield of CO2 was 1.10 g. Say the experiment were carried out in a lab and the amount of CO2 was found to be only 0.78 g. In that case the percent yield is calculated as seen below:
0.787 g --------- × 100% = 71% 1.10 g
The following exercises were written in the order given to help you to develop the skills necessary to master limiting reactant, theoretical yield, and percent yield. You will need a calculator and a periodic table to complete them. You may also use your ions reference sheet.
Introduction to Limiting ReactantN2 + 3H2 2NH3
Nitrogen gas reacts with hydrogen gas to synthesize ammonia gas.
Basic Limiting Reactant Problems5C + 2SO2 CS2 + 4CO
Solid carbon reacts with sulfur dioxide to make carbon disulfide and carbon monoxide.
Limiting Reactant and Percent Yield Problems