In this topic of IBDP Chemistry, we will be discussing all the different types of questions and calculations you are required to know about chemical reactions, including calculating theoretical yields, determining the limiting reagent and converting from various units into moles.

**Theoretical Yield**

In IBDP Chemistry, you will be expected to be able to calculate the amount of products that are formed when given a certain amount of reactants. This is known as calculating the **theoretical yield**. In order to determine the theoretical yield, you will have to follow the steps below.

- Write out the balanced chemical equation for the reaction. (Note: this step is extremely important. In your IBDP Chemistry exams, you may be given the chemical equation, but it may not be balanced, so remember to check this first.)
- Using the information given about the amount of reactants you have (regardless of the units), and convert the amount into moles.
- Once you have obtained this amount in moles, you can determine the amount of moles for the desired product using their molar ratio (this is given by the coeffecients of the balanced equation).
- From there, you may be asked to convert the moles of the product into a different unit (e.g. you may be asked to give the answer in grams).

We will now go through an example of how to calculate theoretical yield.

__ Example 1:__ 10 grams of copper (I) oxide was combined with excess copper (I) sulfide to produce copper and sulfur dioxide. Determine the mass of copper formed from this reaction.

- Our first step would be to write out the balanced chemical equation. We are given the reactants (copper (I) oxide and copper (I) sulfide) and the products (copper and sulfur dioxide). Using this information, we can first write out the reactants and products, then balance the equation.
- The chemical equation is: 2Cu
_{2}O + Cu_{2}S → 6Cu + SO_{2. }If you are unsure of how to balance equations, please click__here__. - Next, we have to calculate the moles of copper (I) oxide using the information on the mass given.
- To do this, we have to first calculate the relative molecular mass of copper (I) oxide.
- Relative molecular mass of Cu2O = (2 x 63.55) + 16 = 143.1gmol
^{-1} - We can then use this to convert to moles.
- Mole of Cu2O = 10g / 143.1gmol
^{-1}= 0.0699mol - Now that we have the number of Cu2O in moles, we can use the molar ratio to determine how many moles of copper will be produced.
- Looking at the chemical equation, we can see that 2 moles of Cu2O will create 6 moles of copper - there is a 1:3 ratio of copper (I) oxide to copper.
- Thus, given that we have 0.0699mol of copper (I) oxide, we will have 3 times the amount of copper formed - the amount of copper formed = 3 x 0.0699mol = 0.210mol copper.
- The questions requires us to give an answer in grams, so we need to convert this value back into grams.
- We can do this by using the relative molar mass of copper - this is written in the periodic table.
- Grams of copper = 63.55gmol
^{-1}x 0.210mol =**13.3g**

**Limiting Reagents**

Now that we know how to determine the theoretical yield of a chemical reaction, we will need to consider something called limiting reagents. In IBDP Chemistry, the **limiting reagent** is defined as the reactant which is completely used up. In other words, it is the reactant with the smallest amount (in mol), relative to the molar ratios.

- The reactant that is used up in the reaction is called the limiting reagent, whereas the reactant that is not used up (i.e. the reactant that will not run out) is called the
**excess reagent**.

By knowing the limiting reagent, we can do further calculations to determine theoretical yield, as well as the amount of other reactants used. Thus, determining the limiting reagent is extremely important, as it gives us the baseline amount of reactants used and products formed.

In IBDP Chemistry exams, you will often be asked to first determine the limiting reagent, and then determine the theoretical yield. We will now go through a worked example.

__ Example 2:__ A student reacted 7.40 x 10

^{-2}g of magnesium ribbon (Mg) to 15cm

^{3 }of 2.00moldm

^{-3}hydrochloric acid (HCl). How many grams of MgCl

_{2}was formed?

- For this question, we are given the amount of both reactant, and we are asked to calculate the mass of magnesium chloride (MgCl
_{2}) formed. - First, as with all questions like this, we need to write out the balanced chemical equation.
- The chemical equation for this reaction is: Mg + 2HCl → MgCl
_{2}+ H_{2} - We will then need to determine the limiting reagent. This was not explicitly asked in the question, but in order to correctly solve the problem, we will have to find the limiting reagent.
- To find the limiting reagent, we should first determine the amount of moles we have for each reactant.
- For magnesium, we are given the mass, so we will need to convert this into moles.
- Mole of Mg = 7.40 x 10
^{-2}g / 24.31gmol^{-1}= 0.00304mol - For hydrochloric acid, we are given the volume and concentration of the solution. To calculate the moles of HCl, we will first need to convert 15cm
^{3}into dm^{3}, and then multiply this value by the concentration (i.e. 2.00moldm^{-3}) - To convert 15cm
^{3}into dm^{3}, we need to divide it by 10^{3}, since 1 dm^{3}= 10^{3}cm^{3}. Therefore, 15cm^{3}= 0.015dm^{3} - Now, we need to multiply 0.015dm
^{3}by 2 to get the number of moles: moles of HCl = 0.015dm^{3}x 2.00moldm^{-3}= 0.03mol - Now that we have calculated the number of moles of each reactant, it is time to determine the limiting reagent.
- From the chemical equation, we can see that the ratio of Mg to HCl is 1:2. Thus, to determine the limiting reagent, we can either multiply the number of moles of Mg by 2, and compare with the number of moles of HCl, or we can divide the number of moles of HCl by 2 and compare with the number of moles of Mg.
- Dividing the moles of HCl by 2, we can see that if we use 0.03mol of HCl, we will need 0.03/2 = 0.015mol of Mg. This is more than the amount of Mg we have, thus magnesium is the limiting reagent.
- Now, we can determine the amount of MgCl
_{2}formed. - We know that we have 0.00304mol of Mg being used, and there is a 1:1 ratio of Mg to MgCl
_{2}. This means that 0.00304mol of MgCl2 will be formed. - chemistryConverting this to grams: mass of MgCl
_{2}= (24.31 + (2 x 35.45)) x 0.00304mol =**0.289g**

**Converting to moles from different units**

As you see in the examples above, many questions in IBDP Chemistry exams related to stoichiometric relationships require you to be able to convert from various units to moles and back. The diagram below gives you a quick overview has to how to make these conversions.

One trick to be able to learn these faster is to look at the units given.

- For example, for conversions between mass and moles, you know you will need to use molar mass, which has the units g/mol. This gives you a hint as to how you can convert, as based on the units, you know that molar mass = mass (g) / moles (mol). You can then substitute the known values into this equation, and rearrange to find the missing value.
- The same can be applied to concentration and volume of solutions - concentration = moles (mol) / volume (dm
^{3}). Substitue known values and then rearrange to find the unknown value.

This is the end of this topic.