posted by Shelby .
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.