between 0.500g of aluminum and excess oxygen in the atmosphere,which is a limiting reactant
To determine the limiting reactant between 0.500g of aluminum and excess oxygen in the atmosphere, we need to compare the amounts of each reactant and determine which one will be completely consumed in the reaction.
First, we need to convert the mass of aluminum to moles. The molar mass of aluminum is 26.98 g/mol.
0.500g Aluminum * (1 mol/26.98g) = 0.01852 mol Aluminum
Next, we need to consider the balanced chemical equation for the reaction between aluminum and oxygen:
4Al + 3O2 → 2Al2O3
From the balanced equation, we can see that it requires 4 moles of aluminum for every 3 moles of oxygen to form 2 moles of aluminum oxide.
So, we can calculate the amount of oxygen required to react with 0.01852 moles of aluminum:
0.01852 mol Aluminum * (3 mol Oxygen / 4 mol Aluminum) = 0.01389 mol Oxygen
Since there is excess oxygen in the atmosphere, which means there is more oxygen available than needed, the limiting reactant is the aluminum.
Therefore, in this reaction, 0.500g of aluminum is the limiting reactant.
To determine which reactant is the limiting reactant, we need to compare the amount of each reactant involved in the reaction.
1. Start by writing and balancing the chemical equation for the reaction:
4 Al + 3 O2 -> 2 Al2O3
2. Convert the mass of aluminum (0.500g) to moles by using the molar mass of aluminum:
Molar mass of aluminum (Al) = 26.98 g/mol
Moles of Al = 0.500g / 26.98 g/mol = 0.0185 mol
3. From the balanced equation, we can see that the stoichiometric ratio between aluminum (Al) and oxygen (O2) is 4:3. This means that for every 4 moles of aluminum, we need 3 moles of oxygen to react completely.
4. Calculate the moles of oxygen required for the reaction using the stoichiometric ratio:
Moles of O2 required = (3/4) x 0.0185 mol = 0.0139 mol
5. Since the reaction takes place in the atmosphere, there is an abundant amount of oxygen available. Therefore, the amount of oxygen is in excess, and aluminum is the limiting reactant.
Conclusion: Aluminum (Al) is the limiting reactant in the reaction between 0.500g of aluminum and excess oxygen in the atmosphere.
To determine which reactant is the limiting reactant, we need to compare the number of moles of each reactant and their respective stoichiometric coefficients in the balanced equation.
Let's start by calculating the number of moles for each reactant.
1. Aluminum (Al):
The molar mass of aluminum (Al) is 26.98 g/mol.
Given: Mass of aluminum = 0.500 g
To calculate the number of moles, we use the formula:
Number of moles = Mass / Molar mass
Number of moles of aluminum = 0.500 g / 26.98 g/mol
Number of moles of aluminum = 0.01853 mol
2. Oxygen (O2):
Oxygen exists as a diatomic molecule (O2), so we need to consider that when calculating the molar mass.
The molar mass of oxygen (O2) is 32.00 g/mol.
Since oxygen is in excess, we can consider it to have an unlimited supply, so we don't need to calculate the moles.
Now that we have both the moles of aluminum and the moles of oxygen, let's compare their stoichiometric coefficients in the balanced equation.
The balanced equation for the reaction between aluminum and oxygen is:
4 Al + 3 O2 → 2 Al2O3
From the balanced equation, we can see that the stoichiometric ratio of aluminum to oxygen is 4:3.
Considering this ratio, for every 4 moles of aluminum, we need 3 moles of oxygen. Hence, the ratio is 4:3.
Comparing the moles of aluminum (0.01853 mol) to the moles of oxygen (unlimited), we can see that the amount of oxygen is much larger.
Therefore, the limiting reactant is aluminum because it is present in a lower quantity compared to the amount required for the reaction.