Reacting Masses

Balance equations using reacting masses

Firstly, the masses of different reactants can be used in many different ways. Using these masses, we are going to explore what exactly we can find out using them. In order to do this, we need to be able to balance an equation and this gives us the stoichiometry of a reaction.

If we have the masses of the reactants, we can identify the masses of products, as well as the moles. But first let’s look at how we can balance an equation through this example.

\[N_2 + H_2 \rightarrow NH_3\]

On the right-hand side we have 2 nitrogen and 2 hydrogens, and on the left there is one nitrogen and 3 hydrogens. So, we already know the equation is not balanced.

So, first, let's fix the amount of nitrogen. If we double the amount of nitrogen on the left-hand side, we will have 2 nitrogens on both sides. We can do this by placing a 2 in front of the N on the left side, to look like this:

\[N_2 + H_2 \rightarrow 2NH_3\]

This now shows as we have 2 nitrogens and 2 hydrogens on the right side and 2 nitrogens and 6 hydrogens on the left. To equal the amount of hydrogen we need 4 more so we can multiply it by 3, by adding a 3 in front of hydrogen on the right side to look like this:

\[N_2 + 3H_2 \rightarrow 2NH_3\]

Now we can see there are 2 nitrogens and both sides and 6 hydrogens on both sides. Therefore, the equation is balanced.

Reacting Masses formula

In this section of the article, we are going to explore the different equations required to determine different problems.

We will explore 4 equations:

• Moles and mass
• Percentage yield
• Moles and volume
• Volume of gasses

Moles and mass

This particular equation is most likely one that you will use throughout your entire career as a chemist. It explores the relationship between moles, mass and molecular mass.

The equation is:

\[\text{mass of substance (g)} = \text{number of moles (mol)} \times \text{molecular mass (\( \mathrm{g}\,\mathrm{mol}^{-1} \))}\]

We can rearrange the equation to calculate moles or molecular mass, and two of the three factors are given to work out what we need.

Percentage yield

Percentage yield allows us to identify how much of a product should be made in theory, by using this formula:

\[\text{yield precentage} = \text{actual yield} \div \text{theoretical yield} \times 100 \]

The actual yield is the amount (moles) of the product that we have actually produced. The theoretical yield is the amount (moles) of the product that we have calculated to produce, we can use the stoichiometry of an equation to calculate this. It is key to remember that both the actual yield and theoretical yield must be in moles.

We divide the actual yield with the theoretical yield, then multiply by 100 to determine the percentage yield.

Moles and volume

We have already explored the relationship between moles, mass and molecular mass. The following equation is different as we explore the relationship between moles, concentration and volume:

\( \text {number of moles (mol)} \) = \( \text {concentrations} \, \mathrm{(mol~dm^{-3})} \times \text {volume} \, \mathrm ({dm^3}) \)

We can see in this equation that it can be used to find the volume, concentration or moles of a solution, as long as we have two of the factors, of course. It is important to note here that volume is \(dm^3\), so we must ensure that any values that we use are in the agreement with each other. This should also be noted when using a concentration value.

Volume of gasses

The last equation we will be exploring is the relationship between volume, molar gas volume and amount of gas in moles. This equation can only be used in conditions known as standard temperature and pressure. The standard temperature is 25 ℃ and the standard pressure is 1 atmosphere. In these conditions, the molar gas volume is fixed at 24.5 dm³. We can explore the equation below:

\( \text {Volume of gas} \, (dm^3) = \text {amount of gas} \, (moles) \times \text {molar gas volume} (24.5~dm^3) \)

Like other equations involving volume, we must ensure the figures we use are in the correct units.

Reacting Masses steps

We have seen the formulae to calculate moles, mass, percentage yield, and volume. Now, we will understand the basic steps involved to calculate some unknown parameter in a problem using the given data.

For example, if you want to calculate the mass of a compound produced when two substances react, one of whose masses is given, we will follow the following steps to calculate the unknown mass of a product.

1. Write the balanced equation for the reaction using the information given in the question.
2. Calculate the molecular masses of all the reactants and products involved in the reaction.
3. Calculate the number of moles from the molecular mass and the reacting mass(given in the question).
4. Based on the stoichiometric coefficient, calculate the molar ratio of the reactant and product( X:Y = X moles of a reactant reacts to produce Y moles of a product, which means the ratio is X:Y).
5. Calculate the mass of the product reacted using the product's molecular mass and the number of moles calculated from step-4.

This is just one example of how to design steps for solving problems based on the information given. We will see some practice questions in the following section where we will explore steps involved in solving problems where other parameters such a moles, volume and percentage yield are involved. We will explain the steps simultaneously through some examples.

Reacting Masses Practice Questions

Now we are going to explore how we can exactly, step-by-step, calculate 4 different types of problems, which are:

• Calculating mass
• Percentage yield
• Excess and limiting reagent
• Volume of a solution
• Volume of gasses

Percentage yield

Firstly for percentage yield we need the equation which is:

Percentage yield = actual yield ÷ theoretical yield x 100

Now let us go through an example:

Excess and limiting reagent

To work out what is a limiting reagent or what is an excess reagent, we need to know what they are exactly. An excess reagent is a reagent that is in excess during a reaction and will be left over in a reaction. A limiting reagent is all used up during the reaction and the reaction stops once the limiting reactant is used up. Thus, the time taken for the reaction to complete is determin

To work these two factors out, we need the moles of both reagents and the stoichiometry.

Let us go through an example.

Volume of solutions

In this section, we are going to explore how we can calculate the volume of a solution. As previously mentioned, we need to ensure all figures used are in the correct units.

We will go through an example to consolidate our learning:

Volume of gasses

The last example we are going to go through will be using the volume of gasses equation. This is usually one of the easiest, as we do not need to find out too much information ourselves. Therefore, let us go through our last example.