How many mL of .715 M HCI is required to neutralize 1.25grams of sodium carbonate? (Producing carbonic acid)

To determine the volume of 0.715 M HCl needed to neutralize 1.25 grams of sodium carbonate, we need to follow these steps:

1. Calculate the number of moles of sodium carbonate (Na2CO3):
- We know the molar mass of sodium carbonate (Na2CO3) is:
- Na: 22.99 g/mol
- C: 12.01 g/mol
- O: 16.00 g/mol
- Therefore, the molar mass of Na2CO3 is 22.99 * 2 + 12.01 + 16.00 * 3 = 105.99 g/mol.
- To find the number of moles, we divide the given mass (1.25 grams) by the molar mass: 1.25 g / 105.99 g/mol = 0.0118 moles.

2. Determine the balanced chemical equation and stoichiometry between HCl and Na2CO3:
- The balanced equation for the reaction between HCl and Na2CO3 is:
2 HCl + Na2CO3 → 2 NaCl + H2O + CO2
- According to the equation, 2 moles of HCl react with 1 mole of Na2CO3.

3. Calculate the number of moles of HCl required:
- Since the stoichiometry between HCl and Na2CO3 is 2:1, we need twice as many moles of HCl as Na2CO3. Thus, we have: 0.0118 moles Na2CO3 * 2 moles HCl/1 mole Na2CO3 = 0.0236 moles HCl.

4. Finally, calculate the volume of 0.715 M HCl solution needed:
- The concentration of the HCl solution is 0.715 M, which means there are 0.715 moles of HCl present in 1 liter (1000 mL) of solution.
- Therefore, to find the volume (V) of HCl required, we divide the number of moles of HCl needed (0.0236 moles) by the molarity of the HCl solution (0.715 M):
V = 0.0236 moles HCl / 0.715 moles/Liter = 0.033 mL

So, approximately 0.033 mL of 0.715 M HCl solution is required to neutralize 1.25 grams of sodium carbonate.

Actually it produces H2O and CO2 although you might argue that it produces H2CO3 a femtosecond before it breaks down to H2O and CO2.

Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 = 1.25/molar mass Na2CO3.
mols HCl = twice that (look at the coefficients in the balanced equation).
M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.