Question

determind the actual and percent yield of NaC2H3O2

opt mmand 3. Determine the amount of excess reactant that remained in the flask. You will need to convert 0.062 mol of HC2H302 into grams to determined how much HC2H302 you started with, so you can subtract how much you used to determine the amount left in the flask. ь.obral icad 60.0619Haita z?72 3 Halb0e 28 9Nablen lholhalles: Ihnol Hahahd9Hang./6392 4. Calculate the actual yield of NaC2Hs02 by using the weight of the empty flask and subtracting it from the weight of the flask after the reaction. actual yrel 5. Using the actual yield of NaC2HsO2 from # 3 and theoretical yield from #2, calculate the percent yield of NaC2H302 recovered in this lab. X/00-59% 6, was your percent yield of NaC2H302 100%? what factors might have caused any error you found? Explain, citing specific examples. Data Table 20 Score Step Substance Mass g NaHCOs g NaHCO3 Needed 0.025 mol NaHCOOD (calculate) mass Weighed2? g NaHCO g NaHCOamount22 g of Flask alone (empty) 3 500 mL Flask Cooled Flask with NaC2H302 834 g of Flask with NaCaHaO2 Questions: 1. Wite the balanced chemical equation for this reaction. (See introduction tor reactants and products.) 2. Calculate the limiting and excess reactants using actual mass of NaHCOs and moles of HC2HaO2. n o

Answer 1

Solution :-

Q1) Balanced reaction equation

NaHCO3+CH3COOH --- > NaC2H3O2 + H2CO3

Q2) Finding the limiting reactant

Moles of CH3COOH = 0.062 mol

Lets calculate the moles of NaHCO3

Moles = mass / molar mass

          = 2.28 g / 84.006 g per mol

          = 0.0271 mol

Mole ratio of the reactant is 1 : 1 , therefore , reactant with less number of moles is the limiting reactant

Therefore the NaHCO3 is the limiting reactant and CH3COOH is the excess reactant

Lets calculate the theoretical yield using the moles of limiting reactant

(0.0271 mol NaHCO3 * 1 mol NaC2H3O2/1 mol NaHCO3)*(82.03 g / 1 mol NaC2H3O2) = 2.22 g NaC2H3O2

Q3)

Amount of excess reactant remained in the flask

Since the mole ratio is 1 : 1 therefore moles of CH3COOH reacted are 0.0271 moles

Moles of CH3COOH remained = starting moles – reacted moles

                                                   = 0.062 mol – 0.0271 mol

                                                   = 0.0349 mol

Mass= moles x molar mass

Mass of CH3COOH = 0.0349 mol * 60.05 g per mol

                                 = 2.10 g

So the amount of excess CH3COOH remain = 2.10 g

Q4) Actual yield of NaHCO3 = (mass of flask + product) – (mass of empty flask)

                                                 = 168.34 g – 166.1 g

                                                 = 2.24 g

Q5) Percent yield = [Actual yield / theoretical yield]*100%

                              =[ 2.24 g / 2.22 g]*100%

                             = 101 %

Q6) The percent yield is more than 100 % this means there is minor errors possible in the experiment

The possible error may be that the mass recorded is may having little higher value that actual product .

There may be some reactant present in the flask which gives the higher mass of the product than actual mass which lead to the higher percentage yield.