Solid manganese dioxide reacts with aqueous hydrochloric acid to form manganese(II) chloride, water, and chlorine gas. If 261 g of MnO ₂ is reacted with 292 g of HCl , the volume of chlorine gas produced at STP is x 10⁻¹ L . The value of x is ____. (Given: Molar masses in g mol ⁻¹ : Mn = 55 , O = 16 , H = 1 , Cl = 35.5 . Molar volume of gas at STP = 22.7 L )

Step-by-step solution

The balanced chemical equation for the reaction is: MnO ₂ + 4 HCl MnCl ₂ + 2 H ₂ O + Cl ₂ First, calculate the number of moles of each reactant. Molar mass of MnO ₂ = 55 + (2 16) = 87 g mol ⁻¹ Moles of MnO ₂ = 261 87 = 3 mol Molar mass of HCl = 1 + 35.5 = 36.5 g mol ⁻¹ Moles of HCl = 292 36.5 = 8 mol From the balanced equation, 1 mol of MnO ₂ requires 4 mol of HCl . So, 3 mol of MnO ₂ would require 12 mol of HCl . Since only 8 mol of HCl is available, HCl is the limiting reagent. The amount of Cl ₂ produced depends