To calculate the number of moles of aluminum used in the combustion reaction, we need to determine the molar mass of aluminum (Al) and convert the given mass of aluminum to moles.
The molar mass of aluminum is 26.98 g/mol.
To convert grams of aluminum to moles, use the formula:
Number of moles = Mass / Molar mass
Number of moles = 0.500 g / 26.98 g/mol
Number of moles ≈ 0.01852 mol
Therefore, approximately 0.01852 moles of aluminum were used in the combustion reaction.
how many moles of aluminium were used in the combustion reaction of 0.500grams of aluminium and oxygen in an atmosphere
3 answers
which is a limiting reactant between 0.500grams of aluminium and excess oxygen in an atmosphere
To determine which reactant is the limiting reactant, we need to calculate the moles of aluminum and oxygen and compare the ratios of the reactants involved in the balanced equation.
Given:
Mass of aluminum = 0.500 grams
Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of oxygen (O₂) = 32.00 g/mol (assuming the diatomic form)
Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum
Number of moles of aluminum = 0.500 g / 26.98 g/mol ≈ 0.01852 mol
Since the mole ratio between aluminum and oxygen in the balanced equation is 4:3, namely:
4 Al + 3 O₂ → 2 Al₂O₃
The theoretical amount of oxygen required to react with all the aluminum can be calculated using the mole ratio:
Number of moles of oxygen needed = (3 mol O₂ / 4 mol Al) * Number of moles of aluminum
Number of moles of oxygen needed = (3/4) * 0.01852 mol ≈ 0.01389 mol
Since oxygen is in excess, and the calculated number of moles (0.01389 mol) needed is lower than the actual amount present in the atmosphere, it means that aluminum is the limiting reactant.
Given:
Mass of aluminum = 0.500 grams
Molar mass of aluminum (Al) = 26.98 g/mol
Molar mass of oxygen (O₂) = 32.00 g/mol (assuming the diatomic form)
Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum
Number of moles of aluminum = 0.500 g / 26.98 g/mol ≈ 0.01852 mol
Since the mole ratio between aluminum and oxygen in the balanced equation is 4:3, namely:
4 Al + 3 O₂ → 2 Al₂O₃
The theoretical amount of oxygen required to react with all the aluminum can be calculated using the mole ratio:
Number of moles of oxygen needed = (3 mol O₂ / 4 mol Al) * Number of moles of aluminum
Number of moles of oxygen needed = (3/4) * 0.01852 mol ≈ 0.01389 mol
Since oxygen is in excess, and the calculated number of moles (0.01389 mol) needed is lower than the actual amount present in the atmosphere, it means that aluminum is the limiting reactant.