You might have noticed a fizz when you mix baking soda and vinegar. We all know that that fizz (effervescence) is due to the evolution of carbon dioxide. But, have you wondered, what amount of baking soda and vinegar produces carbon-di-oxide. What would be the amount of \(CO_2\) produced? What happens if you take one of the reactants (soda and vinegar) in excess? In this article, we will explore all these aspects.
- We will look at how to balance equations using reacting masses
- We will explore the formulas in regard to; moles and mass, percentage yield, moles and volume and volume and gasses.
- Then, we will go over the steps to solving problems when reacting masses are given.
- Furthermore, we will also go through a step-by-step process of determining percentage yield, excess and limiting reagents and volume of gasses using some practice questions alongside.
- We will deep dive into an alternative approach to solving problems when reacting masses of a reactant are given and asked to determine the mass of the product.
- Finally, we will go through some examples to consolidate our knowledge.
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.
Stoichiometry: The relationship between the reactants and products, which can be used to determine data.
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.
- Write the balanced equation for the reaction using the information given in the question.
- Calculate the molecular masses of all the reactants and products involved in the reaction.
- Calculate the number of moles from the molecular mass and the reacting mass(given in the question).
- 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).
- 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
Calculating mass
First, let us establish what equation we need:
\[\text{mass of substance (g)} = \text{number of moles (mol)} \times \text{molecular mass (\( \mathrm{g}\,\mathrm{mol}^{-1} \))}\]
Now let’s go through an example:
Calculate how much mass of calcium oxide can be produced by completely burning 6g of calcium in oxygen.
1. We need to have a balance equation:
\[ 2Ca(s) + O_2(g) \rightarrow 2CaO(s)\]
We can see when comparing both sides of the equation that there are two calcium on each side and two oxygens on each side.
2. Now we need the molecular mass of the reactants and products:
\[Ca = 40\]
\[2 \times O = 32\]
\[CaO = 56\]
3. We need to now calculate the moles of calcium used. We can use the mass provided and the molecular mass of calcium.
The equation we are going to use is:
\( \text {moles (mol)} = \text {mass} \, \mathrm(g) \times \text {Volume} \, \mathrm {dm}^3 \)
We want moles so that means we need to rearrange the equation:
\( \text{Moles of Calcium (moles)} = \text{mass of calcium (g)} \div \text{molecular mass of calcium} \, \mathrm {g~ mol^{-1}} \)
So
\[\text{Moles of Calcium (moles)} = 6 (g) ÷ 40 (g mol^{-1}) = 0.15 moles\]
4. Now, we need to look at the stoichiometry of the equation to identify the molar ratio
\[ 2Ca(s) + O_2(g) \rightarrow 2CaO(s)\]
We can see the calcium and calcium oxide both have a 2 in front of them, this means they have a 2:2 ratio, which can be simplified to 1:1 Therefore, whatever the moles of calcium is, the moles of calcium oxide will be the same. So, the moles of calcium oxide is also 0.15 moles.
5. Finally, we need to rearrange our equation again to find the mass of calcium oxide
\[\text{Mass of calcium oxide (g)} = \text{moles of calcium oxide (moles)} \times \text{molecular mass of calcium oxide (g mol^{-1})}\]
\[\text{Mass of calcium oxide (g)} = 0.15 (moles) \times 56 (gmol^{-1})\]
\[\text{Mass of calcium oxide (g)} = 8.4 g\]
So, our final answer is that 6g of calcium completely burnt with oxygen produced 8.4g of calcium oxide.
Reacting Masses Calculations
There is an alternate approach to the above question where you don't have to go through the moles. Let us take the same example as above and see how we can solve it alternatively.
Calculate how much mass of calcium oxide can be produced by completely burning 6g of calcium in oxygen.
Solution:
Underline/ highlight the data given and data to be calculated in the question.
1. Write the balanced equation
\( 2Ca + O_2 \rightarrow 2CaO \)
2. Calculate the molar mass of calcium and molecular mass of calcium oxide as we have been asked to find the reacting mass of calcium oxide, and we have been given the reacting mass of calcium.
Molar mass of \(Ca = 40\) = In the above equation, the stoichiometric coefficient is 2. Therefore,
\( 2~ Ca = 2 \times 40 = 80g\)
Molecular mass of \(CaO = 56\). According to the equation, \( 2 CaO = 2 \times 56 = 112 g\)
Therefore, according to the equation, we can say that 80 g of Ca produces 112 g of CaO. Then, 6 g of Ca can produce Xg of CaO.
$$ 80 g \rightarrow 112 g~of~CaO $$
$$ 6 g \rightarrow X g~of~CaO $$
On cross multiplication, we get,
\( X = \dfrac {6 \times 112} {80} \)
On simplifying the fraction,
\( X = 8.4 g \)
Therefore, 8.4 g of Calcium oxide is produced when 6 g of Ca is burnt in oxygen.
Thus, we can get the answer through this short-cut method. But, it is advisable to go through the moles if you have time in your exam.
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:
6g of calcium reacts with oxygen to produce calcium oxide. The actual yield was 6.7g. Calculate the percentage yield.
1. We need to have a balanced equation:
\[ 2Ca(s) + O_2(g) \rightarrow 2CaO(s)\]
We can see when comparing both sides of the equation that there are two calciums on each side and two oxygens on each side.
2. Now we need the molecular mass of the reactants and products:
\[Ca = 40\]
\[2 \times O = 32\]
\[CaO = 56\]
3. We need to now calculate the moles of calcium used. We can use the mass provided and the molecular mass of calcium.
The equation we are going to use is:
\[\text{mass of substance (g)} = \text{number of moles (mol)} \times \text{molecular mass (\( \mathrm{g}\,\mathrm{mol}^{-1} \))}\]
We want moles, so that means we need to rearrange the equation:
\(\text{Moles of Calcium (moles)} =\text{mass of calcium (g)} ÷ \text{ molecular mass of calcium} (g~mol^{-1}) \)
So
\[\text{Moles of Calcium (moles)} = 6 (g) \div 40 (g mol^{-1}) = 0.15 moles\]
4. Now we need to look at the stoichiometry of the equation to identify the molar ratio
\[ 2Ca(s) + O_2(g) \rightarrow 2CaO(s)\]
We can see the calcium and calcium oxide both have a 2 in front of them, this means they have a 2:2 ratio which can be simplified to 1:1. Therefore, whatever the moles of calcium is, the moles of calcium oxide will be the same. So, the moles of calcium oxide is also 0.15 moles, this is also the theoretical amount in moles.
5. Now we need to identify what the moles of the actual yield were.
We can do this using the same equation as before, then plugging in the mass of the actual yield produced and the molecular mass of calcium oxide.
\[\text{Moles of Calcium oxide (moles)} = \text{mass of calcium oxide (g)} \div \text{ molecular mass of calcium oxide (g mol^{-1})}\]
\[\text{Moles of Calcium oxide (moles)} = 6.7 (g) ÷ 56 (g mol^{-1})\]
\[\text{Moles of Calcium oxide (moles)} = 0.12 moles\]
0.12 moles is our actual yield.
6. Now we have the moles of actual yield and theoretical yield, we plug the figures into the equation:
\( \text {Yield percentage} = 0.12 \div 0.15 \times 100 \)
Percentage yield = 80%
So, the percentage yield for this reaction is 80%.
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.
9.7g of sodium reacts with 8.5g of sulphur to produce Na2S. State which reagent is limiting and which is in excess.
Write the balanced equation.
\( 2Na + S \rightarrow Na_2S \)
1. First, calculate the moles of each reactant
\[\text{Moles of sodium (moles)} = \text{mass of sodium (g)} ÷ \text{molecular mass of sodium} (g~mol^{-1})\]
\[\text{Moles of sodium (moles)} = 9.7 g ÷ 23 (g mol^{-1})\]
\[\text{Moles of sodium (moles)} = 0.42 moles\]
\[\text{Moles of sulphur (moles)} = \text{mass of sulphur (g)} ÷ \text{molecular mass of sulphur} (g mol^{-1}) \]
\[\text{Moles of sulphur (moles)} = 8.5 g ÷ 32 (g mol^{-1})\]
Moles of sulphur (moles) = 0.27 moles
2. Now we need a balanced equation to identify the stoichiometry
\[ 2Na + S \rightarrow Na_2S\]
3. We can see on both sides there are 2 sodiums and 1 sulphur, so the equation is balanced.
4. Finally, we can identify the limiting and excess reagent.
We can see for 0.4 moles of sodium to react we need 0.2 moles of sulphur. As there is 0.27 moles of sulphur, it means sulphur is in excess, making it the excess reagent and sodium the limiting reagent.
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:
25 cm3 of 0.1 mol*dm-3 concentration of hydrochloric acid was used to neutralise 20 cm3 of sodium carbonate solution of unknown concentration. Calculate the concentration of sodium carbonate.
1. First, we need to calculate the moles of hydrochloric acid
\[\text{Moles of hydrochloric acid (moles)} = \text{concentration of hydrochloric acid} (mol~dm^{-3}) \times \text{volume of hydrochloric acid} (dm^3) \]
We can see here that the volume of hydrochloric acid is in cm3 to convert this to dm3 we need to divide it by 1000, which equals to = 0.025 dm3 it is now in the correct units to input into the equation.
\[\text{Moles of hydrochloric acid (moles)} = 0.025 dm^3 \times 0.1 mol~dm^{-3}\]
Moles of hydrochloric acid (moles) = 0.0025 moles
2. We now need the balanced equation of this reaction to calculate the stoichiometry.
\[ 2HCl + Na_2CO_3 \rightarrow 2NaCl + H_2O + CO_2\]
We can see that the stoichiometric mole ratio of sodium carbonate to hydrochloric acid is 1:2. So, 0.0025 moles of hydrochloric acid is required to react with 0.00125 moles of sodium carbonate.
3. Now we can calculate the concentration using the moles of sodium carbonate and the volume we were provided with.
We were provided the volume of the calcium carbonate solution in cm3, so we need to do what we did previously by dividing the volume of 20cm3 by 1000 to get the correct units.
\[20 ÷ 1000 = 0.02 dm^3\]
4. Now we can use this figure to calculate the concentration.
Concentration of sodium carbonate solution (mol dm-3) = moles of sodium carbonate (moles) ÷ volume of sodium carbonate (dm-3)
\[\text {Concentrations of sodium carbonate solution} (mol dm^{-3}) = 0.00125 (moles) ÷ 0.02 dm^3\]
\[\text{Concentrations of sodium carbonate solution} (mol dm^{-3}) = 0.0625 mol dm^{-3}\]
So the concentration of sodium carbonate solution (mol dm-3) = 0.0625 mol dm-3
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.
Calculate the volume of 3 moles of hydrogen under standard conditions.
We know the number of moles of hydrogen, and we know the molar gas volume (24.5 dm3), so to calculate the volume of the gas, we use the following equation:
\[\text{Volume of hydrogen} (dm^3) = \text{amount of hydrogen (moles)} \times \text{molar gas volume} (24.5 dm^3)\]
\[\text{Volume of hydrogen} (dm^3) = 3 moles x 24.5 dm^3\]
\[\text{Volume of hydrogen} (dm^3) = 73.5 dm^3\]
So, the volume of 3 moles of hydrogen is 73.5 dm3
Reacting Masses - Key takeaways
- Using the masses of different reactants, we can calculate numerous things such as; moles and mass, percentage yield, excessing and limiting reagents, moles and volume and volume of gasses.
- To calculate the mass of a substance using moles and molecular mass, we can use the equation: \[\text{Mass of substance (g)} = \text{number of moles (moles)} \times \text{ molecular mass (g mol^{-1})}\]
- To calculate percentage yield, we use the equation: \[\text{Percentage yield} = \text{actual yield} \div \text{theoretical yield} \times 100\]
- To calculate moles, volume and concentration we can use the equation: \[Moles (moles) = concentration (mol dm^{-3}) x volume (dm^3)\]
- To calculate the volume of gasses, we can use the equation: \[\text{Volume of gas (dm^3)} = \text{amount of gas (moles)} \times \text{ molar gas volume} (24 dm^3)\]
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