Answer:
The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.
Step-by-step explanation:
Given data:
Mass of aluminium = 7.0 g
Mass of oxygen = 15.2 g
Limiting reactant = ?
Solution:
Chemical equation:
4Al + 3O₂ → 2Al₂O₃
Number of moles of Aluminium:
Number of moles = mass/ molar mass
Number of moles = 7.0 g / 27 g/mol
Number of moles = 0.3 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 15.2 g / 32 g/mol
Number of moles = 0.5 mol
Now we will compare the moles of aluminium oxide with oxygen and Al.
O₂ : Al₂O₃
3 : 2
0.5 ; 2/3×0.5 = 0.33 mol
Al : Al₂O₃
4 : 2
0.3 : 2/4×0.3 = 0.15 mol
The number of moles of Al₂O₃ produced by Al are less it will limiting reactant.