Problem: For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.2Al(s) + 3Cl2(g) → 2AlCl3(s)2.5 g Al, 2.5 g Cl2

2 Al(s) + 3 Cl₂(g) → 2 AlCl₃(s)

Calculate the amount of AlCl₃ produced from each reactant and determine the limiting reactant:

mass Al (molar mass) → moles Al(mole-to-mole comparison) → moles AlCl₃(molar mass) → mass AlCl₃

molar mass Al = 26.982 g/mol
molar mass AlCl₃ = 133.332 g/mol

$\mathbf{mass}\mathbf{ }{\mathbf{AlCl}}_{\mathbf{3}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Al}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Al}}}{\mathbf{26}\mathbf{.}\mathbf{982}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Al}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{AlCl}}_{\mathbf{3}}}}{\mathbf{2}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Al}}}\mathbf{\times }\frac{\mathbf{133}\mathbf{.}\mathbf{332}\mathbf{ }\mathbf{g}\mathbf{ }{\mathbf{AlCl}}_{\mathbf{3}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{AlCl}}_{\mathbf{3}}}}$

mass AlCl₃ = 12.35 g

mass Cl₂ (molar mass) → moles Cl₂ (mole-to-mole comparison) → moles AlCl₃ (molar mass) → mass AlCl₃