Answer:
Mass = 124 g
Explanation:
Given data:
Mass of Al = 100 g
Mass of chlorine = 100 g
Mass of aluminum chloride produced = ?
Solution:
Chemical equation:
2Al + 3Cl₂ → 2AlCl₃
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 100 g/ 71 g/mol
Number of moles = 1.4 mol
Number of moles of Al:
Number of moles = mass/molar mass
Number of moles = 100 g/ 27 g/mol
Number of moles = 3.7 mol
Now we will compare the moles of AlCl₃ with Al and Cl₂.
Al : AlCl₃
2 : 2
3.7 : 3.7
Cl₂ : AlCl₃
3 : 2
1.4 : 2/3×1.4 = 0.93
Less number of moles of AlCl₃ are produced by Cl₂ thus it will act as limiting reactant.
Mass of AlCl₃:
Mass = number of moles × molar mass
Mass = 0.93 mol × 133.34 g/mol
Mass = 124 g