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 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?

Short answer

In conclusion, the limiting reactant is NaOH. When 1.85 moles of NaOH and 1.00 mole of CO₂ react, 0.925 moles of Na₂CO₃ can be produced, and 0.075 moles of CO₂ remain after the completion of the reaction.

Step-by-step solution

Identify the given quantities

In this problem, we are given the number of moles of both reactants: - 1.85 moles of NaOH (sodium hydroxide) - 1.00 moles of CO₂ (carbon dioxide)

Calculate the mole ratios for the reactants

Using the balanced chemical equation, we can determine the mole ratio between the reactants. In this case, the balanced reaction is: \[2 \mathrm{NaOH}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l)\] The mole ratios in the balanced equation are 2 moles of NaOH to 1 mole of CO₂, or 2:1.

Determine the limiting reactant

To determine the limiting reactant, we need to compare the mole ratios of each reactant to the balanced equation. Divide the number of moles of each reactant by their corresponding mole ratio: For NaOH: \(\frac{1.85 \, \text{moles}}{2}\) = 0.925 For CO₂: \(\frac{1.00 \, \text{moles}}{1}\) = 1.00 The smaller value corresponds to the limiting reactant. In this case, NaOH is the limiting reactant.

Calculate the moles of Na₂CO₃ produced

Using the moles of the limiting reactant (NaOH) and the stoichiometry of the balanced equation, we can determine the moles of Na₂CO₃ produced: Mole ratio of NaOH to Na₂CO₃ is 2:1. So, moles of Na₂CO₃ produced = 0.925 moles (corresponding to 1.85 moles of NaOH)

Determine the excess reactant and the remaining moles

Since NaOH is the limiting reactant, CO₂ is the excess reactant. To determine the remaining moles of CO₂, calculate the moles of CO₂ consumed by the reaction: Mole ratio of NaOH to CO₂ is 2:1. So, moles of CO₂ consumed = \(\frac{1.85 \, \text{moles} \,\text{of } \text{NaOH}}{2} = 0.925 \, \text{moles}\) Now, subtract the moles of CO₂ consumed from the initial moles of CO₂: Remaining moles of CO₂ = \(1.00 \, \text{moles} - 0.925 \, \text{moles} = 0.075 \, \text{moles}\) In conclusion, NaOH is the limiting reactant, 0.925 moles of Na₂CO₃ can be produced, and 0.075 moles of CO₂ remain after the completion of the reaction.

Key concepts

Mole Ratio

When dealing with chemical reactions, the concept of mole ratio is essential. It helps us understand the proportion of reactants and products involved in a reaction. Mole ratio is derived from the coefficients of the balanced chemical equation. By referring to the given problem, let's examine the equation: \[2 \mathrm{NaOH}(s) + \mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l)\]From this equation, we understand that:

• 2 moles of NaOH react with 1 mole of CO₂ to produce 1 mole of Na₂CO₃ and 1 mole of H₂O.
• The mole ratio, in this case, is 2:1 for NaOH to CO₂.

Analyzing the actual amount of reactants, we can predict how much product can form and identify the limiting reactant based on these calculated ratios. The limiting reactant is the one that produces less product when calculated separately with the initial given amounts.

Chemical Equation Balancing

Balancing chemical equations ensures we stay true to the law of conservation of mass, which states that matter cannot be created or destroyed. In a balanced chemical equation, the number of each type of atom is the same on both the reactant and product sides.To balance chemical reactions:

• Start by writing the unbalanced equation.
• Count the number of atoms of each element on both sides.
• Adjust coefficients to get the same number of each type of atom on both sides.

In our given reaction, \[2 \mathrm{NaOH} + \mathrm{CO}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3} + \mathrm{H}_{2} \mathrm{O}\]the equation is balanced. We have an equal number of each type of atom on both sides. This allows us to accurately determine the mole ratio and use this to find limiting and excess reactants.

Stoichiometry

Stoichiometry involves using the principles of balanced equations to predict the quantities of reactants and products involved in a reaction. Once we balance the equation, stoichiometry helps in understanding and solving chemical quantitative problems like the given exercise. Here's how stoichiometry is applied:

• Utilize the balanced equation to find mole ratios which relate quantities of reactants to products.
• Identify the limiting reactant by comparing mole ratios of reactants with the stoichiometric coefficients.
• Calculate amount of product that can be generated from the limiting reactant.
• Determine any remaining excess reactants after the reaction has gone to completion.

In the reaction between NaOH and CO₂, stoichiometric calculations helped us find that NaOH is the limiting reactant since it produces the least amount of product (Na₂CO₃). The excess CO₂ then can be calculated by determining the leftover amount after the reaction completes, showcasing the practical applications of stoichiometry.