Problem 75 - Chemistry The Central Science

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

The limiting reactant is NaOH. A total of 0.925 moles of Na2CO3 can be produced, and 0.075 moles of CO2 remain after the completion of the reaction.

Step 1: Understand the Balanced Chemical Equation

The given balanced chemical equation is: \[2 NaOH (s) + CO_{2} (g) \longrightarrow Na_{2} CO_{3} (s) + H_{2}O (l)\] The coefficients in the balanced equation give the mole ratios between reactants and products. In this case: - 2 moles of NaOH react with 1 mole of CO2 to produce 1 mole of Na2CO3 and 1 mole of H2O.

Step 2: Analyze the Given Quantities of Reactants

We are given the following quantities of reactants: - 1.85 moles of NaOH - 1.00 moles of CO2

Step 3: Determine the Limiting Reactant and the Excess Reactant

We need to compare the given quantities of reactants with the stoichiometric ratios from the balanced equation. Calculate the amount of NaOH needed to react completely with 1.00 moles of CO2: \(1.00 \, moles \, CO_{2} \times \frac{2 \, moles \, NaOH}{1 \, mole \, CO_{2}} = 2.00 \, moles \, NaOH\) However, we only have 1.85 moles of NaOH available. Therefore, the limiting reactant will be NaOH, as there is not enough NaOH to completely react with the given amount of CO2. The excess reactant will be CO2.

Step 4: Calculate the Number of Moles of Na2CO3 Produced

As NaOH is the limiting reactant, the number of moles of Na2CO3 produced can be found using its stoichiometric ratio with NaOH: \(1.85 \, moles \, NaOH \times \frac{1 \, mole \, Na_{2}CO_{3}}{2 \, moles \, NaOH} = 0.925 \, moles \, Na_{2}CO_{3}\) So, 0.925 moles of Na2CO3 can be produced.

Step 5: Calculate the Number of Moles of Excess Reactant Remaining

To find the remaining amount of CO2, we need to see how much of it reacted with the limiting reactant, NaOH: \(1.85 \, moles \, NaOH \times \frac{1 \, mole \, CO_{2}}{2 \, moles \, NaOH} = 0.925 \, moles \, CO_{2} \, reacted\) Now subtract the reacted amount of CO2 from the initial amount: \(1.00 \, mole \, CO_{2} - 0.925 \, moles \, CO_{2} = 0.075\, moles \, CO_{2}\, remaining\) Therefore, 0.075 moles of CO2 remain after the reaction is complete. In summary, the limiting reactant is NaOH, 0.925 moles of Na2CO3 can be produced, and 0.075 moles of CO2 remain after the completion of the reaction.

Key Concepts

Balanced Chemical Equation

Understanding a balanced chemical equation is crucial when dealing with chemical reactions. Here, the equation for the reaction between sodium hydroxide (NaOH) and carbon dioxide (CO\(_2\)) is:\[2 \mathrm{NaOH}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l)\]In this equation, the coefficients (the numbers in front of each compound) tell us the ratio in which these chemicals react. This means:

• 2 moles of NaOH react with 1 mole of CO\(_2\).
• They produce 1 mole of Na\(_2\)CO\(_3\) and 1 mole of H\(_2\)O.

These coefficients are key in predicting how much of each reactant is needed and how much product is formed. By ensuring each side of the equation has the same number of atoms for each element, the equation respects the law of conservation of mass. This balance offers a roadmap to follow when carrying out stoichiometric calculations.

Stoichiometry

Stoichiometry is the study of the quantitative relationships or ratios between reactants and products in a chemical reaction. It relies upon the balanced chemical equation as a guide.To solve exercises involving stoichiometry:

• First, determine the number of moles of each reactant available.
• Use the mole ratios provided by the balanced equation to find out how much of each substance is consumed or produced.

In our example, with 1.85 moles of NaOH and 1.00 mole of CO\(_2\), stoichiometry helps us identify NaOH as the limiting reactant. The balanced equation indicates we need 2 moles of NaOH for every 1 mole of CO\(_2\). Given only 1.85 moles of NaOH are present, it cannot entirely react with the 1.00 mole of CO\(_2\). Therefore, the maximum amount of product, Na\(_2\)CO\(_3\), based on stoichiometric calculations, is limited by the availability of NaOH. This results in 0.925 moles of Na\(_2\)CO\(_3\) being formed.

Excess Reactant

In a chemical reaction, the excess reactant is what remains after the limiting reactant is fully consumed. It's crucial to identify it to understand which reactant will determine the extent of the reaction.For our reaction:

• NaOH is the limiting reactant, as we have only 1.85 moles while needing 2 moles to completely react with 1 mole of CO\(_2\).
• CO\(_2\) becomes the excess reactant.

After determining that only 0.925 moles of CO\(_2\) will react with the available NaOH, we subtract this from the initial amount of CO\(_2\) (1.00 mole). This leaves us with 0.075 moles of CO\(_2\) as the excess reactant.
Recognizing the excess reactant can be practically useful, for example, when deciding how much of other chemicals may need to be added in processes that require complete reactions.