Na₂CO₃ (s) + 2HCL (ag) → 2NaCl (ag) + CO₂ + H₂O (l)

If 4.25g of sodium carbonate is reacted completely with excess hydrochloric acid as 3M HCl, how many mL of HCl would be consumed? Note: Assume 3M HCl contains 3.00 moles of HCl per liter of solution.

mols Na2CO3 = grams/molar mass

mols HCl = 2x mols Na2CO3 (from the coefficients).
M HCl = mols HC/L HCl.
You know M HCl and mols HCl, solve for L and convert to mL.
By the way, you don't need to assume that 3M HCl contains 3 mols HCl/L soln. Why? Because that IS the definition of M. It DOES contain 3 mols HCl/L soln.

To find out how many mL of HCl would be consumed in the reaction, we need to use stoichiometry and the given information. Here's how you can solve the problem:

1. Determine the molar mass of sodium carbonate (Na₂CO₃):
Na: 22.99 g/mol * 2 = 45.98 g/mol
C: 12.01 g/mol
O: 16.00 g/mol * 3 = 48.00 g/mol

Total molar mass of Na₂CO₃ = 106.99 g/mol

2. Convert the given mass of sodium carbonate to moles:
moles of Na₂CO₃ = mass / molar mass
moles of Na₂CO₃ = 4.25 g / 106.99 g/mol = 0.0397 mol

3. Use the balanced equation to find the mole ratio between Na₂CO₃ and HCl.
From the balanced equation, we know that 1 mol of Na₂CO₃ reacts with 2 mol of HCl.

4. Calculate the moles of HCl required to react with the given amount of Na₂CO₃:
moles of HCl = moles of Na₂CO₃ * (2 mol HCl / 1 mol Na₂CO₃)
moles of HCl = 0.0397 mol * (2 mol HCl / 1 mol Na₂CO₃) = 0.0794 mol

5. Convert moles of HCl to volume (in liters) using the given HCl concentration:
volume of HCl (in L) = moles of HCl / concentration of HCl
volume of HCl (in L) = 0.0794 mol / 3 M = 0.0265 L

6. Convert volume from liters to milliliters:
volume of HCl (in mL) = 0.0265 L * 1000 mL/L = 26.5 mL

Therefore, 26.5 mL of HCl would be consumed in the reaction.