Final answer:
HCl is used up first since we only have 1.322 moles available, which will limit the reaction. Hence, 23.44 g of Cl2 will be produced when 0.86 moles of MnO2 and 48.2 g of HCl react.
Step-by-step explanation:
To determine which reagent is used up first in the reaction MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O, we need to perform a stoichiometric calculation. First, let's find out how many moles of HCl we have. The molar mass of HCl is approximately 36.46 g/mol. With 48.2 g of HCl, we have 48.2 g / 36.46 g/mol = 1.322 mol of HCl.
According to the balanced equation, 4 moles of HCl are required for every 1 mole of MnO₂. Therefore, 0.86 moles of MnO₂ would need 0.86 * 4 = 3.44 moles of HCl. Since we have only 1.322 moles of HCl available, HCl is the limiting reactant.
Since 4 moles of HCl produce 1 mole of Cl₂, with 1.322 moles of HCl, we can produce 1.322 / 4 = 0.3305 moles of Cl₂. The molar mass of Cl₂ is approximately 70.90 g/mol, so 0.3305 moles of Cl₂ will have a mass of 0.3305 * 70.90 g/mol = 23.44 g of Cl₂.