Limiting Reactant In LiOH And CO2 Reaction A Comprehensive Guide

Jul 17, 2025 by ADMIN 65 views

Hey guys! Let's dive into a classic chemistry problem involving limiting reactants. This is a crucial concept in stoichiometry, which helps us understand how much product we can make from a given set of reactants. We've got a fun reaction on our hands: carbon dioxide (CO2) reacting with lithium hydroxide (LiOH) to produce lithium carbonate (Li2CO3) and water (H2O). We'll figure out which reactant limits the amount of product formed.

The Reaction

First, let’s take a look at the balanced chemical equation:

$CO_2 + 2 LiOH \rightarrow Li_2CO_3 + H_2O$

This equation tells us that one mole of carbon dioxide reacts with two moles of lithium hydroxide to produce one mole of lithium carbonate and one mole of water. It's like a recipe – you need the right proportions to get the perfect cake! We are given that $1.000 imes 10^3$ g of LiOH is combined with $8.80 imes 10^2$ g of $CO_2$ . The experiment yields $3.25 imes 10^2$ g of $H_2O$ . The burning question is: what's the limiting reactant in this scenario?

Understanding Limiting Reactants

Before we jump into calculations, let's clarify what a limiting reactant actually is. Imagine you're making sandwiches. You have 20 slices of bread and 15 slices of cheese. You can only make 10 sandwiches because you'll run out of bread first. Bread is your limiting ingredient, it limits the amount of sandwiches you can make. Similarly, in a chemical reaction, the limiting reactant is the one that is completely consumed first, thereby dictating the maximum amount of product that can be formed. The other reactant(s) are in excess – there's more than enough of them to react.

Identifying the limiting reactant is super important because it tells us the theoretical yield of the reaction, which is the maximum amount of product we can possibly obtain. To find the limiting reactant, we need to compare the moles of reactants we have with the stoichiometry of the balanced equation. It's all about ratios!

Step-by-Step Identification of the Limiting Reactant

Okay, let's break down how to find the limiting reactant in our reaction. We'll go through it step-by-step to make sure we've got it down.

Step 1: Convert Grams to Moles

We can't directly compare grams of reactants because each molecule has a different mass. We need to convert the masses of LiOH and CO2 to moles using their respective molar masses. Remember, the molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).

Molar mass of LiOH: Lithium (Li) has a molar mass of approximately 6.94 g/mol, Oxygen (O) is about 16.00 g/mol, and Hydrogen (H) is roughly 1.01 g/mol. Adding these up, we get: 6. 94 + 16.00 + 1.01 = 23.95 g/mol. So, one mole of LiOH weighs about 23.95 grams.
Molar mass of CO2: Carbon (C) has a molar mass of about 12.01 g/mol, and Oxygen (O) is 16.00 g/mol. Since there are two oxygen atoms in CO2, we have: 12.01 + (2 * 16.00) = 44.01 g/mol. Thus, one mole of CO2 weighs around 44.01 grams.

Now, let's convert the given masses to moles:

Moles of LiOH: We have 1.000 x 10^3 g of LiOH. To convert this to moles, we divide by the molar mass: (1000 g) / (23.95 g/mol) ≈ 41.75 moles of LiOH.
Moles of CO2: We have 8.80 x 10^2 g of CO2. Converting to moles: (880 g) / (44.01 g/mol) ≈ 20.00 moles of CO2.

Step 2: Determine the Mole Ratio

Now we need to look at the balanced equation again:

$CO_2 + 2 LiOH \rightarrow Li_2CO_3 + H_2O$

This tells us that 1 mole of CO2 reacts with 2 moles of LiOH. This is our mole ratio – the key to figuring out the limiting reactant. We need to compare the ratio of moles we have to this ideal ratio.

Step 3: Identify the Limiting Reactant

There are a couple of ways to do this. Here’s one method: divide the moles of each reactant by its coefficient in the balanced equation:

For LiOH: 41.75 moles / 2 (coefficient) ≈ 20.88
For CO2: 20.00 moles / 1 (coefficient) = 20.00

The reactant with the smaller result is the limiting reactant. In this case, CO2 (20.00) is smaller than LiOH (20.88), so CO2 is the limiting reactant. This means we'll run out of CO2 before we run out of LiOH.

Another way to think about it is: Do we have enough LiOH to react with all the CO2? We have 20.00 moles of CO2. According to the balanced equation, we need twice as many moles of LiOH, which would be 2 * 20.00 = 40.00 moles of LiOH. We have 41.75 moles of LiOH, which is more than 40.00 moles, so LiOH is in excess. Conversely, do we have enough CO2 to react with all the LiOH? We have 41.75 moles of LiOH. We need half as many moles of CO2, which would be 41.75 / 2 = 20.88 moles of CO2. We only have 20.00 moles of CO2, which is less than 20.88 moles, confirming that CO2 is the limiting reactant.

Why is this Important?

Identifying the limiting reactant isn't just a theoretical exercise. It has practical implications in chemistry and many other fields. For instance, in industrial chemical processes, knowing the limiting reactant allows chemists to optimize reactions, maximize product yield, and minimize waste. It also helps in accurately predicting the amount of product that can be formed, which is crucial for cost-effectiveness and resource management.

Calculating Theoretical Yield

Since we know CO2 is the limiting reactant, we can calculate the theoretical yield of the products, like water (H2O). The balanced equation tells us that 1 mole of CO2 produces 1 mole of H2O. We started with 20.00 moles of CO2, so theoretically, we should produce 20.00 moles of H2O.

To convert this to grams, we use the molar mass of H2O. Hydrogen is approximately 1.01 g/mol, and Oxygen is 16.00 g/mol, so H2O has a molar mass of (2 * 1.01) + 16.00 = 18.02 g/mol.

Theoretical yield of H2O = 20.00 moles * 18.02 g/mol ≈ 360.4 grams

This means, in an ideal scenario, the reaction should produce 360.4 grams of water. However, in the experiment, only 325 grams of water were produced. This brings us to the concept of percent yield.

Percent Yield

The percent yield compares the actual yield (the amount of product obtained in the experiment) to the theoretical yield (the maximum possible amount of product). It’s a measure of the reaction's efficiency. The formula for percent yield is:

Percent Yield = (Actual Yield / Theoretical Yield) * 100%

In our case:

Actual Yield = 325 grams of H2O
Theoretical Yield = 360.4 grams of H2O

Percent Yield = (325 g / 360.4 g) * 100% ≈ 90.18%

So, the percent yield of the reaction is approximately 90.18%. This indicates that the reaction was quite efficient, as we obtained a high percentage of the maximum possible product.

Factors Affecting Yield

It's rare to achieve a 100% yield in a chemical reaction. Several factors can influence the actual yield:

Incomplete Reactions: Some reactions don't go to completion, meaning not all reactants convert to products.
Side Reactions: Reactants might participate in other, unintended reactions, leading to byproducts and reducing the yield of the desired product.
Loss During Transfer: During the experimental process, some product might be lost during transfer between containers or during purification steps.
Experimental Errors: Human errors, such as inaccurate measurements or spills, can also affect the yield.

Conclusion

So, to wrap it up, in the reaction between LiOH and CO2, we determined that CO2 is the limiting reactant. We figured this out by converting the masses of reactants to moles, comparing the mole ratio to the balanced equation, and identifying which reactant would run out first. Understanding limiting reactants is fundamental in chemistry, guys, as it helps us predict the maximum amount of product we can form and optimize reactions for better yields. We also touched on theoretical yield, percent yield, and factors that can affect the efficiency of a reaction. Keep these concepts in mind, and you'll ace your stoichiometry problems!

Limiting Reactant Explained CO2 and LiOH Reaction Chemistry