PEREAKSI PEMBATAS

In a chemical reaction, when the substances react, not all of the reactions are used to form the product. Well, in this case, there are several possibilities that can happen.

For example, let's say there is a reaction between the substances A and B resulting in C and D. When A and B react, there are several possibilities that can happen. The first possibility is that the two reactions, namely A and B, both have a reaction that forms C and D.

The second possibility is that both reactions, both A and B, do not end in reaction. Or, in other words, after the reaction is finished, both are still left. And the third possibility is that one of the reactions is finished. For example, A is finished reacting and B is still left, or vice versa.

In this case, the reaction that ends first is called the limiting reaction. This limiting reaction is very important for you to understand because it is the basis for many more continuous chemical calculations. For example, we need a limiting reaction to determine the remaining or formed mass

Determine the molar volume or concentration of matter after a reaction happens mathematically. The question is, how do we determine the boundary reaction through calculation? Let's learn together. The boundary reaction in a reaction can be determined through the following steps. First, we determine the initial mole of each reaction first.

Secondly, we do a simple calculation, namely we divide the mole of each reaction by the coefficient of each. The reaction with the smallest division is the limiting reaction. To understand more, let's continue with the examples of questions.

The first example, 1 mol sodium hydroxide solution NaOH is reacted with 1 mol sulfate acid H2SO4 according to the following reaction: 2 NaOH + H2SO4 = Na2SO4 + 2 H2O Determine A, the limiting reaction, B, the remaining reaction.

As we have learned before, we need to know the initial moles of each reaction to determine the limiting reaction. Where for this question, the initial moles of the two reactions, in this case NaOH and H2SO4, have been known, namely as much as 1 mole.

Next, we divide the mol NaOH and H2SO4 with their respective coefficients. For the reaction coefficients, you can see the equation in the question, where for NaOH the coefficient is 2 and for H2SO4 the coefficient is 1.

So for NaOH, the mole of NaOH is divided by the coefficient equal to 1 mole divided by 2, the result is 0.5 moles. Then for H2SO4, the mole of H2SO4 is divided by the coefficient equal to 1 mole divided by 1, the result is 1 mole.

From this distribution result, we can see that the result of the NAOH division is smaller than H2SO4, so the limiting reaction for this issue is NAOH and the remaining reaction is H2SO4.

Let's continue to the second question. 3.2 g of methane with the formula of CH4 molecule is reacted with 16 g of oxygen according to the following reaction equation. CH4 + 2O2 produces CO2 + 2H2O.

If the relative mass of CH4 is 16 g/mol and the relative mass of oxygen is 32 g/mol, determine the first mol of the initial reaction, the second reaction of the limiter, and the third of the remaining reaction mass.

Well, in this question, the reaction is CH4 and O2, which means for question A, what we have to determine is the initial mole of CH4 and O2. In previous videos, we have discussed the concept of moles, where a molecule's mole can be calculated by the formula that its mass is divided by its relative molecular mass.

For CH4, the question is known that its mass is 3.2 grams and the relative molecule or MR is 16 grams per mole, so the CH4 mole is equal to 3.2 grams divided by 16 grams per mole. The result is 0.2 moles.

Next for oxygen, the mass is known to be 16 grams and the mass of the relative molecule or MR is 32 grams per mole. So the mole of oxygen is equal to 16 grams divided by 32 grams per mole. The result is 0.5 moles. We continue to the second question, we must determine the reaction of the limit.

As we have discussed, we divide the mole of each reaction first with the coefficients of each. For CH4, we have calculated the moles at point A, which is 0.2 moles, and the coefficients are 1.

then the CH4 mol divided by its coefficient is equal to 0.2 mol divided by 1, the result is 0.2 mol. Then for oxygen, based on the previous calculation results, the mol is 0.5 mol, and from the reaction equation, we can see that the coefficient is 2.

so the mole of oxygen divided by the coefficient is equal to 0.5 moles divided by 2, the result is 0.25 moles. From these two calculations, we can see that the result of the division of CH4 is smaller than O2, so the limiting reaction for this issue is CH4.

Next for point C, we are asked to determine the remaining reactions. To make it easier, we can make a table like this, there is a common reaction ratio, then the initial mole, the mole used to react, and the remaining mole of the reaction and the mole of the product that is formed.

For the initial moles, we have calculated that for CH4 the initial mole is 0.2 mol and for the oxygen, the initial mole is 0.5 mol. For CO2 and H2O, we just empty the initial moles because the product is not formed at the beginning.

Then, according to point B that we have worked on, the limiting reaction was CH4. Remember that this limiting reaction is the reaction that is exhausted first.

It means that the number of mol of the limiting reaction used to react is the same as the original mol. So for CH4, because the original mol is 0.2 mol, then what is used to react is also 0.2 mol.

Moreover, for O2 that reacts and CO2 and H2O that are formed, we determine by matching the CH4 mol, namely by comparing the coefficients to the CH4 coefficients.

Therefore, for a question like this, you have to make sure that the reaction is even and see if the coefficient is correct or not. Well, for this question, the reaction is even, and for those of you who are still confused about the equation of reactions,

or determine the coefficient, you can open the video link in the video description about the interaction and how to calculate it. Okay, now let's take a look at the coefficient comparison. For CH4, the coefficient is 1, while for oxygen or O2, the coefficient is 2.

It can be seen that the coefficient of O2 is 2 times the coefficient of CH4, so the mol is also 2 times. If CH4 reacts as much as 0.2 mol, then for O2 which reacts as much as 2 times 0.2 mol, which is 0.4 mol.

Next for CO2, the coefficient is 1, the same as CH4, so that the number is also the same as the reaction CH4, which is 0.2 mol. Even for H2O, the coefficient is 2, so that the number is 2 times 0.2 mol, which is 0.4 mol.

In determining the final or residual condition for reaction, namely CH4 and O2, obtained from the initial mole is reduced by the reaction mole.

For CH4, the initial mole was 0.2 mol, reduced by the reaction, which is 0.2 mol, the result is 0 mol. Which means, in the final condition, CH4 is not left. This is in accordance with the concept of limiting reaction, where limiting reaction does end first compared to other substances.

Moreover, for O2, the initial mole is 0.5 mol and the reaction is 0.4 mol, so at the end, there is 0.1 mol left. Next for the product, namely CO2 and H2O, we just need to drop it like this.

So, in this reaction, CO2 is 0.2 mol and H2O is 0.4 mol. So far we can see that the remaining reaction is O2, which means the remaining reaction time that we will determine is O2.

Now we remember again that mol is equal to mass divided by the mass of the relative molecule, then to find the O2 mass, we move the relative mass of the molecule so that the O2 mass is equal to the mol times the relative mass of the molecule.

For the mol, we already got 0.1 mol. And for the relative molecular mass, it is already known in the question, which is 32 grams per mol. Then the O2 mass is equal to 0.1 mol multiplied by 32 grams per mol.

So