Answer:
5.83 mol.
Explanation:
2Al + 3Ag₂S → 6Ag + Al₂S₃,
It is clear that 2 mol of Al react with 3 mol of Ag₂S to produce 1 mol of Ag and 1 mol of Al₂S₃.
Al reacts with Ag₂S with (2: 3) molar ratio.
So, 2.27 mol of Al reacts completely with 3.4 mol of Ag₂S with (2: 3) molar ratio.
The reamining moles of excess reactant "Al" = 8.1 mol - 2.27 mol = 5.83 mol.