Question

One important factor in any chemical synthesis is the actual quantity of desired product obtained compared to the theoretical amount predicted on the basis of the stoichiometry of the reaction. The ratio of the mass of product obtained to the theoretical quantity, expressed as a percentage, is referred to as the "percent yield" or more simply the "yield". For example: if we react HCl with excess NaOH one of the products will be NaCl. If we assume that all of the Cl in HCl ends up as NaCl we know that each mole of HCl consumed should produce a mole of NaCl product. Suppose that in a particular reaction 10.9 grams of HCl (36.5 g/mol) react with excess NaOH and 19 grams of NaCl (58.5 g/mol) are isolated from the reaction mixture by crystallization. What is the percent yield of NaCl in the experiement?

Answer 1

Answer: the percent yield of NaCl in the experiement is 109%.

Step-by-step explanation:

1) Each mole of HCl consumed should produce a mole of NaCl product

⇒ 1 mol HCl / 1 mol NaCl

2) Reactant: 10.9 grams of HCl (36.5 g/mol)

• Convert grams to moles: moles = mass in grams / molar mass

• moles = 10.9 g / 36.5 g/mol = 0.299 mol HCl

3) Theoretical (stoichiometric) yield of NaCl

• Set the proportion: 1 mol HCl / 1 mol NaCl = 0.299 mol HCl / x

⇒ x = 0.299 mol NaCl (should be obtained)

4) Actual yield

• 19 grams of NaCl (58.5 g/mol)

• Convert to moles: moles = mass in grams / molar mass

• moles = 19 g / 58.5 g/mol = 0.325 mol NaCl

5) Percent yield:

• Equation: percent yield = (actual yiel / theoretical yield) × 100

• Percent yiedl = 109% ← answer

Note: it is not normal to obtain percent yields greater than 100%. When this happens, you must think in an experimental error or that the sample was contaminated with some substance that yield to an increase of the measured product.