Problem 77 - Chemistry The Central Science In Si Units

Question

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

Step-by-Step Solution

Answer

NaOH is the limiting reactant, producing 0.925 mol \(\mathrm{Na}_{2}\mathrm{CO}_{3}\); 0.075 mol \(\mathrm{CO}_{2}\) remains.

Step 1: Write the Balanced Equation

The balanced chemical equation is already provided in the problem: \[2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l)\] This equation will guide us to determine the stoichiometric ratios of the reactants and products.

Step 2: Determine Stoichiometric Ratios

From the balanced equation, 2 moles of \(\mathrm{NaOH}\) react with 1 mole of \(\mathrm{CO}_{2}\) to produce 1 mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). This gives us the ratio \[\frac{2\text{ mol NaOH}}{1\text{ mol CO}_2} = 2\text{ NaOH : 1 CO}_2 \]

Step 3: Calculate the Mole Ratio from the Given Quantities

We are given that there are 1.85 mol of \(\mathrm{NaOH}\) and 1.00 mol of \(\mathrm{CO}_{2}\). Calculate the minimum moles required for \(\mathrm{NaOH}\) to react completely with \(\mathrm{CO}_{2}\):\[1.00 \text{ mol CO}_2 \times \frac{2 \text{ mol NaOH}}{1 \text{ mol CO}_2} = 2.00 \text{ mol NaOH}\] Since only 1.85 mol of \(\mathrm{NaOH}\) is available and 2.00 mol is required, \(\mathrm{NaOH}\) is the limiting reactant.

Step 4: Calculate Moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) Produced

From the balanced equation, 2 moles of \(\mathrm{NaOH}\) produce 1 mole of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\). Therefore, \[1.85 \text{ mol NaOH} \times \frac{1 \text{ mol } \mathrm{Na}_{2}\mathrm{CO}_{3}}{2 \text{ mol NaOH}} = 0.925 \text{ mol } \mathrm{Na}_{2}\mathrm{CO}_{3}\] So, 0.925 moles of \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) can be produced.

Step 5: Calculate the Remaining Moles of \(\mathrm{CO}_{2}\)

Since \(\mathrm{NaOH}\) is the limiting reactant, calculate the remaining moles of \(\mathrm{CO}_{2}\) after the reaction:\[1.85 \text{ mol NaOH} \times \frac{1 \text{ mol CO}_2}{2 \text{ mol NaOH}} = 0.925 \text{ mol CO}_2\] Thus, \(1.00 \text{ mol CO}_2 - 0.925 \text{ mol CO}_2 = 0.075 \text{ mol CO}_2\) remaining.

Key Concepts

Stoichiometry

Understanding stoichiometry is like unlocking the secret code to how substances react in chemistry. It involves using a balanced chemical equation to find the relationships between different substances in a reaction. These relationships are usually expressed as ratios based on the coefficients from the balanced equation.
In the sodium hydroxide (\(\mathrm{NaOH}\) ) and carbon dioxide (\(\mathrm{CO}_2\) ) reaction, stoichiometry helps us establish that 2 moles of \(\mathrm{NaOH}\) reacts with 1 mole of \(\mathrm{CO}_2\) to produce 1 mole of sodium carbonate (\(\mathrm{Na}_2\mathrm{CO}_3\)). By following these stoichiometric ratios, we can predict the quantities of products and reactants involved.
Stoichiometry is essential because it helps chemists:

• Determine the perfect amount of reactants required to fully react without wasting any materials
• Calculate the yield of a reaction, which tells us how much product can be produced
• Identify the limiting reactant, which is the substance that is completely consumed during the reaction and limits the amount of product formed

By mastering stoichiometry, students can solve complex chemical equations with confidence.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into new substances. In this process, old bonds break, and new bonds form, usually accompanied by energy changes.
The reaction of sodium hydroxide with carbon dioxide is an example of a synthesis reaction, where two or more simple substances combine to form a more complex compound. In our equation:\[ 2 \mathrm{NaOH}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l) \]The reactants are changing into products: sodium carbonate and water. Understanding the types of reactions and the conditions necessary for these reactions to occur is crucial in predicting how different chemicals will behave.
More specifically, each reaction type, like combustion, synthesis, decomposition, and replacement, has a unique pattern that can help us understand the chemical world better. This reaction showcases how atmospheric CO₂ can react with substances on Earth to form new materials, revealing the dynamic nature of chemical changes.

Mole Calculations

In chemistry, the mole is a fundamental unit of measure that helps in quantifying atoms, molecules, or other chemical entities. Understanding mole calculations is vital for determining the reactants and products in a given reaction.
When carrying out mole calculations, chemists often:

• Use Avogadro's number, \( 6.022 \times 10^{23} \), to convert between moles and atoms or molecules
• Apply balanced chemical equations to find out the mole ratios of reactants and products
• Identify the limiting reactant to predict the amount of product that can be formed

In our specific case, the mole calculation shows us the exact amounts of \(\mathrm{NaOH}\) \ and \(\mathrm{CO}_2\) \ needed, and it helps to determine that \(\mathrm{NaOH}\) \ is the limiting reactant.
This means that \(\mathrm{NaOH}\) \ will run out first, stopping the reaction from producing more product at 0.925 moles of \(\mathrm{Na}_2\mathrm{CO}_3\)\, while some of the \(\mathrm{CO}_2\) \ remains unused. Mastering mole calculations allows chemists to precisely control reactions and efficiently produce desired substances.