2 Al(s) + 3 Cl_{2}(g) → 2 AlCl_{3}(s)

Calculate moles Cl_{2} using mole-to-mole comparison:

molar mass aluminum (Al) = 26.982 g/mol

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

**moles Cl _{2} = 0.4453 mol Cl_{2}**

Calculate volume Cl_{2} using the ideal gas equation:

Aluminum reacts with chlorine gas to form aluminum chloride.

2 Al(s) + 3 Cl_{2}(g) → 2 AlCl_{3}(s)

What minimum volume of chlorine gas (at 298 K and 251 mmHg ) is required to completely react with 8.01 g of aluminum?

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