An impure sample of manganese dioxide ( MnO ₂ ) weighing 17.4 g and having 50 % purity by mass is reacted with 500 mL of 1.0 M HCl solution. The reaction is given below: MnO ₂( s ) + 4 HCl ( aq ) MnCl ₂( aq ) + Cl ₂( g ) + 2 H ₂ O ( l ) What mass of MnCl ₂ is produced in this reaction? (Given: Molar mass of Mn , O , H and Cl are 55, 16, 1 and 35.5 g mol ⁻¹ , respectively)

Step-by-step solution

Mass of pure MnO ₂ = 17.4 50 100 = 8.7 g Molar mass of MnO ₂ = 55 + 2(16) = 87 g mol ⁻¹ Moles of pure MnO ₂ = 8.7 87 = 0.1 mol Moles of HCl = 1.0 500 1000 = 0.5 mol From the balanced chemical equation, 1 mole of MnO ₂ reacts with 4 moles of HCl . Dividing moles by stoichiometric coefficients to find the limiting reagent: For MnO ₂: 0.1 1 = 0.1 For HCl : 0.5 4 = 0.125 Since 0.1 Moles of MnCl ₂ formed = Moles of MnO ₂ = 0.1 mol Molar mass of MnCl ₂ = 55 + 2(35.5) = 126 g mol ⁻¹ Mass of MnCl ₂ produced = 0.1 126 = 12.