If 15.00 mL of a 1.00 M HCl solution is mixed with 25.00 mL of a 0.250 M NaOH solution in a process that produces 273 mg of NaCl, what is the percent yield?

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Quizplus · Accepted Answer

The percent yield can be calculated by comparing the actual yield (the mass of NaCl produced) to the theoretical yield (the mass of NaCl that should be produced based on stoichiometry). First, calculate the moles of HCl and NaOH:- Moles of HCl = 1.00 M * 0.015 L = 0.015 mol- Moles of NaOH = 0.250 M * 0.025 L = 0.00625 molSince HCl and NaOH react in a 1:1 molar ratio to produce NaCl and water, the limiting reactant is NaOH (it has fewer moles). Therefore, the theoretical yield of NaCl is based on the moles of NaOH.- Moles of NaCl theoretically produced = 0.00625 molThe molar mass of NaCl is approximately 58.44 g/mol, so the theoretical mass of NaCl produced is:- Theoretical mass = 0.00625 mol * 58.44 g/mol = 0.36525 g = 365.25 mgGiven that the actual yield is 273 mg, the percent yield is:- Percent yield = (actual yield / theoretical yield) * 100 = (273 mg / 365.25 mg) * 100 ≈ 74.8%

If 1500 ML of a 1