Answer
AlBr₃ is the limiting reactant
Explanation
Mass of AlBr₃ = 23.45 g
Mole of AlBr₃ = Mass/molar mass = 23.46 g/266.69 g/mol = 0.08797 mol
Mass of Na₂O = 34.57 g
So mole of Na₂O = mass/molar mass = 34.57 g/61.9789 g/mol = 0.5578 mol
Mole ratio of AlBr₃ to Na₂O = (0.08797/0.08797) : (0.05578/0.08797) = 1:6
To know the limiting reactant, write a chemical equation for the reaction and compare the reactant mole ratio in the equation with the mole ratio above.
2AlBr₃ + 3 Na₂O -----> Al₂O₃ + 6NaBr
Mole ratio in the equation = 2:3
Comparing the mole ratio, AlBr₃ will be the first to be completely consumed.
Therefore, AlBr₃ is the limiting reactant
