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 reagent 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?

Short answer

The limiting reactant is NaOH. The amount of Na2CO3 produced is \(0.925\,mol\), and the remaining amount of the excess reactant CO2 is \(0.075\,mol\).

Step-by-step solution

Identify the given information

We are given the following information: - Initial amount of NaOH: \(1.85\, mol\) - Initial amount of CO2: \(1.00\, mol\) - Balanced chemical equation: \(2\,NaOH + CO2 \rightarrow Na2CO3 + H2O\)

Determine the limiting reactant

To find the limiting reactant, we need to compare the ratio of moles of the reactants with the stoichiometric ratio from the balanced equation. The stoichiometric ratio of \(NaOH:CO2\) is \(2:1\). Now, let's find the actual ratio by dividing the number of moles of NaOH by the moles of CO2: \[\frac{1.85\,mol\,NaOH}{1.00\,mol\,CO2} = 1.85\] Since the actual ratio (1.85) is less than the stoichiometric ratio (2), this means that NaOH is the limiting reactant.

Calculate the amount of product that can be formed

Using the stoichiometry of the balanced chemical equation, we can calculate the amount of Na2CO3 produced. The stoichiometric ratio of NaOH to Na2CO3 is \(2:1\). Since we have \(1.85\,mol\) of the limiting reactant NaOH, we can determine the moles of Na2CO3 produced as follows: \[1.85\,mol\,NaOH \times \frac{1\,mol\,Na2CO3}{2\,mol\,NaOH} = 0.925\,mol\,Na2CO3\] So, \(0.925\,mol\) of Na2CO3 can be produced.

Calculate the remaining amount of the excess reactant

Since CO2 is the excess reactant, we can find out how much of it remains after the reaction. We can use the stoichiometric ratio between NaOH and CO2, which is \(2:1\). From the given information, \(1.85\,mol\) of NaOH reacted. Therefore, the moles of CO2 that reacted are: \[1.85\,mol\,NaOH \times \frac{1\,mol\,CO2}{2\,mol\,NaOH} = 0.925\,mol\,CO2\] Now, subtract the moles of CO2 that reacted from the initial amount of CO2: \[1.00\,mol\,CO2 - 0.925\,mol\,CO2 = 0.075\,mol\,CO2\] So, \(0.075\,mol\) of CO2 remain after the completion of the reaction. To sum up: - The limiting reactant is NaOH. - The amount of Na2CO3 produced is \(0.925\,mol\). - The remaining amount of the excess reactant CO2 is \(0.075\,mol\).