Final answer:
In the given reaction between sodium hydroxide and carbon dioxide, sodium hydroxide is the limiting reactant. 0.75 moles of sodium carbonate would form. 0.05 moles of excess carbon dioxide remain after the reaction is complete.
Step-by-step explanation:
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and carbon dioxide (CO2) is:
2NaOH + CO2 → Na2CO3 + H2O
To determine the limiting reactant and the amount of sodium carbonate formed, we need to compare the moles of the reactants to the stoichiometric ratio in the balanced equation.
Given:
Moles of NaOH = 1.5 mol
Moles of CO2 = 0.8 mol
The stoichiometric ratio between NaOH and CO2 is 2:1. Therefore, for every 2 moles of NaOH, 1 mole of CO2 is required. This means that if there were enough CO2 present, we would need twice as many moles of NaOH.
Since we have 1.5 moles of NaOH and 0.8 moles of CO2, it is clear that CO2 is present in excess. In this reaction, NaOH is the limiting reactant. This means that all of the NaOH will react and determine the amount of sodium carbonate formed. Since the stoichiometric ratio is 2:1, we can calculate the moles of sodium carbonate formed:
Moles of Na2CO3 = (1.5 mol NaOH) x (1 mol Na2CO3 / 2 mol NaOH) = 0.75 mol Na2CO3
Regarding the moles of excess reactant left after the reaction is complete, we can calculate the moles of CO2 that did not react:
Moles of excess CO2 = Moles of CO2 initially - Moles of CO2 reacted
= 0.8 mol - 0.75 mol
= 0.05 mol CO2