Problem: Aluminum reacts with chlorine gas to form aluminum chloride.2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)What minimum volume of chlorine gas (at 298 K and 251 mmHg ) is required to completely react with 8.01 g of aluminum?

2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s)

Calculate moles Cl₂ using mole-to-mole comparison:

molar mass aluminum (Al) = 26.982 g/mol

$\mathbf{moles}\mathbf{ }{\mathbf{Cl}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Al}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Al}}}{\mathbf{26}\mathbf{.}\mathbf{982}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Al}}}\mathbf{\times }\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{Cl}}_{\mathbf{2}}}{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Al}}}$

moles Cl₂ = 0.4453 mol Cl₂

Calculate volume Cl₂ using the ideal gas equation: