Limiting Reactant and Percent Yield in a Neutralization
A chemistry student in Kabul reacts 4.0 g of sodium hydroxide (NaOH, molar mass 40 g/mol) with 50.0 mL of 1.0 M hydrochloric acid (HCl). The reaction is NaOH + HCl -> NaCl + H2O. After the reaction, the student evaporates the water and obtains 4.68 g of dry sodium chloride (NaCl, molar mass 58.5 g/mol). The student claims the yield is close to 100 percent, but the instructor asks for a careful analysis of which reactant limits the reaction and what the theoretical and percent yields actually are. Assume the reaction goes to completion and the only solid product is NaCl.
1. Determine the limiting reactant by comparing moles of NaOH and HCl.
Read the worked answer
Moles of NaOH = mass / molar mass = 4.0 / 40 = 0.10 mol. Moles of HCl = M * V = 1.0 mol/L * 0.050 L = 0.050 mol.
The balanced equation shows a 1:1 mole ratio. We have 0.10 mol NaOH but only 0.050 mol HCl. HCl is present in the smaller stoichiometric amount, so HCl is the limiting reactant; NaOH is in excess.
2. Calculate the theoretical yield of NaCl and the percent yield.
Read the worked answer
Since HCl is limiting and the ratio is 1:1, moles of NaCl formed = moles of HCl = 0.050 mol. Theoretical mass of NaCl = 0.050 mol * 58.5 g/mol = 2.925 g.
Percent yield = (actual / theoretical) * 100 = (4.68 / 2.925) * 100 = 160 percent.
This is impossible for a simple reaction, so the student's claim of ~100 percent is wrong: the actual mass exceeds the theoretical maximum, indicating either impure product (e.g., leftover NaOH or water), a measurement error, or that the starting masses were misrecorded.
3. If the student had used 50.0 mL of 2.0 M HCl instead, would the limiting reactant change? Show the calculation.
Read the worked answer
New moles of HCl = 2.0 * 0.050 = 0.10 mol. Moles of NaOH = 0.10 mol.
Now both reactants are present in exactly 0.10 mol, the stoichiometric 1:1 ratio. Neither is limiting in excess - the reaction is stoichiometrically balanced. Theoretical NaCl = 0.10 mol * 58.5 = 5.85 g. If the student still obtained only 4.68 g, the percent yield would be (4.68/5.85)*100 = 80 percent, a plausible result.