Final answer:
To react completely with 0.750 L of 0.400 M Na2CO3, we need 0.300 liters of 2.00 M HCl. This is calculated by determining the moles of Na2CO3, then using the stoichiometry of the balanced equation to find the necessary moles of HCl, and then finding the corresponding volume.
Step-by-step explanation:
To find out what volume of 2.00 M HCl is necessary to react completely with 0.750 L of 0.400 M Na2CO3, we need to first write out the balanced chemical equation for the reaction between hydrochloric acid (HCl) and sodium carbonate (Na2CO3):
2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)
From the balanced equation, we see that the mole ratio of HCl to Na2CO3 is 2:1. Therefore, twice the amount of moles of HCl are needed for each mole of Na2CO3. We can calculate the moles of Na2CO3 using its concentration and volume:
Moles of Na2CO3 = 0.400 M × 0.750 L
= 0.300 moles
Since the mole ratio of HCl to Na2CO3 is 2:1, we will need 2 times the moles of Na2CO3 of HCl:
Moles of HCl needed = 2 × 0.300 moles
= 0.600 moles
Now, we can calculate the volume of 2.00 M HCl needed to provide these moles:
Volume of HCl = Moles of HCl ÷ Molarity of HCl = 0.600 moles ÷ 2.00 M
= 0.300 L
Thus, 0.300 liters of 2.00 M HCl are needed to react completely with 0.750 L of 0.400 M Na2CO3.