How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
157.9 mL (approximately 158 mL) of 0.1 M HCl is required.
Explanation
This problem involves stoichiometry of acid-base reactions. Since the mixture contains equimolar amounts of Na₂CO₃ and NaHCO₃, we first calculate the moles of each component from the total mass. Then, using the balanced chemical equations, we determine the total moles of HCl needed. Na₂CO₃ requires 2 moles of HCl per mole, while NaHCO₃ requires 1 mole of HCl per mole. Finally, using molarity formula, we find the volume of HCl solution required.
Solution Steps
Step 1: Let the mixture contain 'x' moles each of Na₂CO₃ and NaHCO₃ (equimolar amounts).
Step 2: Calculate molar masses: Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 106 g mol⁻¹; Molar mass of NaHCO₃ = 23 + 1 + 12 + 3(16) = 84 g mol⁻¹.
Step 3: Set up mass equation: Total mass = mass of Na₂CO₃ + mass of NaHCO₃. Therefore, 1 g = x × 106 + x × 84 = x × 190. Solving: x = 1/190 moles.
Step 4: Write balanced reactions: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ (2 moles HCl per mole Na₂CO₃); NaHCO₃ + HCl → NaCl + H₂O + CO₂ (1 mole HCl per mole NaHCO₃).
Step 5: Calculate total moles of HCl required: Moles of HCl = 2x + x = 3x = 3 × (1/190) = 3/190 moles.
Step 6: Calculate volume using molarity formula: Volume (L) = Moles/Molarity = (3/190)/0.1 = 3/19 L = 0.1579 L = 157.9 mL.