Asked 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.
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.
Answers
Answered by
DrBob222
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.
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.
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