Consider the following reaction: 2 Al ( s ) + 6 HCl ( aq ) 2 AlCl ₃( aq ) + 3 H ₂( g ) What mass of AlCl ₃ will be formed if 5.4 g of solid aluminum reacts with 200 mL of 1.5 M HCl solution? (Given: Molar mass of Al and Cl are 27 and 35.5 g mol ⁻¹ , respectively)

Step-by-step solution

Moles of Al = 5.4 27 = 0.2 mol Moles of HCl = 1.5 200 1000 = 0.3 mol From the balanced chemical equation, 2 moles of Al react with 6 moles of HCl . Dividing moles by stoichiometric coefficients to find the limiting reagent: For Al : 0.2 2 = 0.1 For HCl : 0.3 6 = 0.05 Since 0.05 Moles of AlCl ₃ formed = 2 6 ( Moles of HCl ) = 1 3 0.3 = 0.1 mol Molar mass of AlCl ₃ = 27 + 3(35.5) = 133.5 g mol ⁻¹ Mass of AlCl ₃ formed = 0.1 133.5 = 13.35 g Answer: 13.35 g