Limiting Reactant Calculator

This free limiting reactant calculator assists you to calculate limiting reactant that goes for finishing during the reaction and makes a limited amount of product.

Let’s go through the kinematics of the limiting reactions in the context below!

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Limiting Reactant Definition:

In the light of chemistry:

“A particular reactant that gets consumed fully when the reaction is completed is known as the limiting reactant”

A chemical reaction can not proceed without this limiting reagent.

Excess Reagent:

At the end of the reaction, the reactant that is left behind in more quantity is known as the excess reagent.

Our best limiting reactant calculator immediately notifies the limiting and excess reagents in a chemical reaction and lets you know about the condition of the reaction.

How To Find Limiting Reactant?

The limiting reagent calculations are performed by considering the following strategies

Balanced Chemical Equations:

One of the highlighted hurdles in limiting reactant problems is the unbalanced chemical equation. You must balance your chemical reaction in order to highlight the limiting reactant. But do not worry as you can access our best chemical equation balancer calculator to equalize the reactant with that of products in no time. Yes, what you think, we get it done for you.

No. of Moles:

The next step involves the determination of the number of moles of reactants and products from the balanced equation that you can instantly by tapping the mole calculator so that you may avoid any wrong calculations during the process.

Mole Ratios:

A very important key factor that leads you to know how to do limiting reactant calculations is the molar ratio of reactants with those of the products formed. You can find these particular ratios by using another mole fraction calculator as well.

Molar Mass:

The molar mass is a key parameter that will help you out in the determination of the limiting reagent. You can find the molecular weightage of any compound in a chemical reaction by using our molar mass calculator. After you do that, you can judge which reactant in the recation is limitiing one.

Determine How Much Product Is Formed:

Here you are required to find what is the partition of each reactant in the formation of the product. If you do not understand the scenario, let this free limiting reactants calculator do this stuff for you.

Identification of Limiting And Excess Reactant:

After all, you are just one step away. Now you need to wfor recognizing that what was the reactant that was fully consumed when reaction got stopped. That reactant will be the limiting one. While all other reactants are considered excess as they still keep sufficient quantity to form product.

How To Find The Limiting Reagent?

Here we will resolve an example that will broaden your mind map regarding the concept. Let’s move on!

Example # 01:

How to find limiting reagent and excess reactant for the limiting reactant equation given below:

$$ N_2 + H_2 →NH_3 $$

Solution:

As the given reaction is not balanced, so its balanced form is as follows:

$$ 1N_2 + 3H_2 → 2NH_3 $$

Finding Mole Ratios:

Here we have:

$$ \text{Mole ratio between N_2 and NH_3} = \frac{\text{1 mol of NH_2}}{\text{2 mol of NH_3}} $$

$$ \text{Mole ratio between H_2 and NH_3} = \frac{\text{3 mol of H_2}}{\text{2 mol of NH_3}} $$

Finding Molar Masses:

$$ \text{Molar Mass of N_2} = 28.02 \frac{g}{mol} $$

$$ \text{Molar Mass of H_2} = 2.016 \frac{g}{mol} $$

$$ \text{Molar Mass of NH_3} = 17.031 \frac{g}{mol} $$

Product Formation From 30g of \(N_2\):

$$ NH_3 \left(grams\right) = \text{30 g of N_2} * \left(\frac{\text{1 mol of N_2}}{\TEXT{28g N_2}}\right) * \left(\frac{\text{2 mol of NH_3}}{\text{1 mol of N_2}}\right) * \left(\frac{\text{17.031g NH_3}}{\text{2 mol of NH_3}}\right) $$

$$ NH_3 \left(grams\right) = 30 * 0.035 * 2 * 8.5155 $$

$$ NH_3 \left(grams\right) = 17.88255 g $$

Product Formation From 90g of \(H_2\):

$$ NH_3 \left(grams\right) = \text{96g of H_2} * \left(\frac{\text{1 mol of H_2}}{2.016g H_2}\right) * \left(\frac{\text{2 moles of NH_3}}{\text{3 moles of H_2}}\right) * \left(\frac{\text{17.031g NH_3}}{\text{2 mol of NH_3}}\right) $$

$$ NH_3 \left(grams\right) = 96 * 0.496 * 0.666 * 8.5155 $$

$$ NH_3 \left(grams\right) = 270.045 g $$

So the limiting reactant is the \(N_2\). This is because it is utilized 100% to form just 17.88255g of the product. While on the other hand, \(H_2\) forms approximately 270.045g of the product which is more than that of \(N_2\). You can also verify these calculations by using our best and only limiting reactant calculator with steps.

Now we will see how to find excess reactant amount that is still remaining and can be used to continue the reaction if more \(N_2\) is added to it.

$$ Grams \left(H_2\right) = \text{270.045g of NH_3} * \left(\frac{\text{2 Moles of NH_3}}{\text{17.031g NH_3}}\right) * \left(\frac{\text{3 moles of H_2}}{\text{2 moles of NH_3}}\right) * \left(\frac{\text{2.016 grams H_2}}{\text{3 mol H_2}}\right) $$

$$ Grams \left(H_2\right) = 270.045 * 0.1174 * 1.5 * 0.672 $$

$$ Grams \left(H_2\right) = 31.956g $$

So we noted that it exactly took 31.956g of \(H_2\) to complete the reaction. The remaining excess amount of this reactant is calculated as follows:

$$ \text{Grams Remaining} = \text{Total Grams – Grams Used} $$

$$ \text{Grams Remaining} = 90g – 31.956g $$

$$ \text{Grams Remaining} = 58.044g $$

So at the moment when the reaction stops, there will be exactly 58.044g of H_2 remaining. Aprt from these complex manual calculations, our free excess reagent calculator takes a couple of moments to display accurate results.

How Limiting Reactant Calculator Works?

This calculator determines which reactant is the limiting one and allows the reaction to stop. Let’s find how!

Input:

• First of all, write down the chemical equation in the designated field or load already stored example equations
• Now tap the calculate button
• You will be displayed with properly balanced equation with coefficients and number of moles given
• Now what you need to do is to enter either number of moles or mass of the reactants in their respective boxes

Output:

The free limiting reagent calculator calculates:

• Limiting reactant involved in the chemical reaction

FAQ’s:

Can there be a limiting reagent if only one reactant is involved in the reaction?

When there is only one reactant in the chemical reaction, a stage comes when the reaction stops suddenly. Here what you need to do is to calculate limiting reagent that will help you sort out the problem. Moreover, you can also take assistance from this best limiting reactant calculator with grams that will let you know how many grams of reactant have been used in the product formation.

Is limiting reactant always the same for given chemical reaction?

Yes, of course it is. A limiting reactant is the one that gets end first to fomr the limited amount of product. So for every specific reaction, it remains the same for that particular one.

How are limiting reactants used in everyday life?

Suppose you are eating bread with delicious jam. The bread gets end soon but the jam is still remaining in the jar. From this you immediately notice that the bread is the limiting reagent while jam iss the excessive one. Many other such examples can be noticed in all around.

Is limiting reagent theoretical yield?

The limiting reactant always produces a liited yield of the product. This limited yield is known as the theoretical yield that you could determine by using free theoretical yield calculator.

Why limiting reactant is so important?

A limiting reactant is the only one that allows the reaction to stop because of no more quantity.

Can a limiting reagent be a gas?

Yes it can be. Many chemical reactions are out tere that have gases involved as reactants and forming various products.

Conclusion:

No doubt without a limiting reactant, you are not capable of determining the minimum possible yield of any chemical reaction. This is why this best limiting reactant calculator with moles allows you to calculate this yield that further provides assistance in keeping the reaction run smoothly.

References:

From the source of Wikipedia: Limiting reagent, Comparison, Comparison of product amounts

From the source of Khan Academy: Limiting reactant, Reaction yields, Stoichiometry

From the source of Lumen Learning: Limiting Reagents, Chemical Reactions