Final answer:
To react completely with 0.750 L of 0.400 M Na₂CO₃, you would need 300 mL of 2.00 M HCl.
Step-by-step explanation:
To determine the volume of 2.00 M HCl needed to react completely with 0.750 L of 0.400 M Na2CO3, we need to use the balanced chemical equation and the concept of stoichiometry. In the chemical equation:
2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
it shows that 2 moles of HCl react with 1 mole of Na2CO3. First, we calculate the moles of Na2CO3 using its concentration:
0.400 M Na2CO3 x 0.750 L = 0.300 moles Na2CO3
Since the ratio of HCl to Na2CO3 is 2:1, we need twice as many moles of HCl:
2 x 0.300 moles HCl = 0.600 moles HCl
Finally, we can use the molarity and the moles of HCl to calculate the volume needed:
0.600 moles HCl / 2.00 M HCl = 0.300 L or 300 mL