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Chapter 3: Problem 77

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

Expert verified
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

01

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)
02

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.
03

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.
04

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)
05

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.

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