Final answer:
The maximum amount of nitrogen dioxide that can be formed is 10.844 grams. The formula for the limiting reagent is N2O, and the amount of the excess reagent (O2) remaining after the reaction is complete is 2.176 grams.
Step-by-step explanation:
To determine the maximum amount of nitrogen dioxide that can be formed, we first need to identify the limiting reagent. The limiting reagent is the reactant that is completely consumed in the reaction and determines the amount of product that can be formed. We can use the given masses of N2O and O2 to calculate the number of moles for each reactant.
First, we convert the masses of N2O (10.4 g) and O2 (9.74 g) to moles by dividing them by their respective molar masses.
N2O: (10.4 g) / (44.02 g/mol) = 0.236 mol
O2: (9.74 g) / (32 g/mol) = 0.304 mol
Since the balanced equation shows a 1:2 mole ratio between N2O and N2O2, we can determine the limiting reagent by comparing the moles of N2O and O2. There is less moles of N2O (0.236 mol) compared to O2 (0.304 mol), so N2O is the limiting reagent.
Now that we know the limiting reagent, we can calculate the theoretical yield of nitrogen dioxide. From the balanced equation, we know that for every 1 mole of N2O, 1 mole of N2O2 is formed. Therefore, the maximum amount of N2O2 that can be formed is equal to the amount of N2O, which is 0.236 mol.
To convert moles to grams, we multiply the number of moles by the molar mass of N2O2:
N2O2: (0.236 mol) * (46.01 g/mol) = 10.844 g
So, the maximum amount of nitrogen dioxide that can be formed is 10.844 grams.
The formula for the limiting reagent is N2O because it is the reactant that limits the formation of the product.
To determine the amount of the excess reagent remaining after the reaction is complete, we need to calculate the amount of O2 that reacted. We can do this by subtracting the moles of O2 that reacted from the total moles of O2 initially present.
From the balanced equation, we know that for every 1 mole of O2, 1 mole of N2O2 is formed. Therefore, the moles of
O2 that reacted is equal to the moles of N2O2 formed, which is 0.236 mol.
The moles of O2 initially present is 0.304 mol. Thus, the moles of O2 remaining is:
O2 remaining: (0.304 mol) - (0.236 mol) = 0.068 mol
To convert moles to grams, we multiply the number of moles by the molar mass of O2:
O2 remaining: (0.068 mol) * (32 g/mol) = 2.176 g
Therefore, the amount of the excess reagent (O2) remaining after the reaction is complete is 2.176 grams.