Final answer:
To dissolve 37.2 g of PbO₂, you need 51.7 mL of 12 M HCl, calculated using stoichiometry by determining the moles of PbO₂ and the corresponding moles of HCl required.
Step-by-step explanation:
Calculating the Volume of Concentrated HCl Required to Dissolve Lead (IV) Oxide
The reaction between lead (IV) oxide and hydrochloric acid can be represented by the following equation: PbO₂(s) + 4HCl(aq) → PbCl₂(s) + Cl₂(g) + 2H₂O(l). To determine the volume of 12 M HCl needed to completely react with 37.2 g of PbO₂, we start by calculating the number of moles of PbO₂.
Step 1: Calculate the molar mass of PbO₂, which is (207.2 g/mol Pb) + (2 × 16.0 g/mol O) = 239.2 g/mol.
Step 2: Calculate the moles of PbO₂ using the mass given: (37.2 g) / (239.2 g/mol) = 0.155 moles of PbO₂.
Step 3: Since 4 moles of HCl react with 1 mole of PbO₂, we multiply the moles of PbO₂ by 4 to get the moles of HCl required: 0.155 moles × 4 = 0.62 moles HCl.
Step 4: Calculate the volume of 12 M HCl needed, knowing that molarity (M) equals moles of solute divided by volume of solution in liters (L): Volume = Moles / Molarity = 0.62 moles / 12 M = 0.0517 L of HCl.
This is equivalent to 51.7 mL of 12 M HCl required to completely dissolve 37.2 g of PbO₂.