Final answer:
To react 125.0 g of NaHCO₃ with 6.00 M HCl, 247.8 mL of HCl is needed, calculated by first determining the moles of NaHCO₃ and then using the molarity to find the volume required.
Step-by-step explanation:
The question asks how many milliliters of 6.00 M HCl are required to completely react with 125.0 g of NaHCO₃. The first step is to calculate the number of moles of NaHCO₃ using its molar mass (84.01 g/mol):
- Number of moles of NaHCO₃ = 125.0 g ÷ 84.01 g/mol = 1.487 moles of NaHCO₃
The balanced chemical equation for the reaction between NaHCO₃ and HCl is:
NaHCO₃(aq) + HCl(aq) → NaCl(aq) + CO₂(g) + H₂O(l)
From the balanced equation, it is clear that the reaction is in a 1:1 molar ratio between NaHCO₃ and HCl. Therefore, 1.487 moles of NaHCO₃ will require 1.487 moles of HCl. Finally, we calculate the volume of HCl required using the molarity (M) definition:
- Volume of HCl = number of moles ÷ molarity = 1.487 moles ÷ 6.00 M
- The volume of HCl = 0.2478 L or 247.8 mL
Thus, 247.8 mL of 6.00 M HCl is needed to react completely with 125.0 g of NaHCO₃.