Question

Activity 10 How Much Product Will My Reaction Yield? 37 6. Carly the chemistry student performs an experiment with vinegar (active ingredient acetic acid, CH, COOH) and baking soda (NaHCO). Her professor asks her to place varying amounts of baking soda and vinegar into scalable plastic sandwich bags, quickly seal the bags, and observe the reaction. The balanced chemical equation for the reaction is given here. NaHCO, (aq) + CH,COOH(ag) - CO,(8) + H,O(l) + NaCH COO(ag) a) Use limiting reagent calculations to fill in the following table. You will need to do your calculations on a separate sheet of paper. Mass of Baking Soda Volume of Vinegar (5.0% Acetic Acid w Moles of CO of Baking Soda Is Limiting Moles of CO If Vinegar is Limiting Reactant Moles of Co Formed ] Limiting 0.25 g 10 mL Baking soda Balks sola. 0.0030 ml 0.17mol 10.009 mol 0.13 mol 0.018ml | 0.084ml 0.0030 MO! 0.0089 mol 0.758 8.0 mL 1.58 5.0 mL Baking soda 10.00lmol b) Which bag will inflate the most will inflate the Is her with the most coa PART II: PERCENT YIELD 7. You are making chocolate chip cookies for a party. You have followed a recipe for making the cookie batter that should produce four dozen cookies. Everything is going well until you forget to set the timer for the last batch and burn 10 of the cookies. You throw the burned cookies away and bring the rest to your party. a) How many cookies did you expect to make from your recipe? 48 cookies b) How many edible cookies were you actually able to bring to your party? 38 cookies c) What percentage of the total cookies you planned on making did you actually bring to the party? 79.16% total cookies brought

Answer 1

100 ml Badanic - Stoichiometric coefficient NaHCO3 + CH₃COOH - CO₂ + H₂00 + CH₂ coo Na Trial A Alaking soda = /2 coz Ner2000 NaHCO₃ 0.25 g MW = 84 glud n = 0.259 = 2.996x10males 0.003 84g/oud 51. W CH3 COOH v=10 ml MW = 60 g/mol = 59 51. " moles = 58 x 10 ml x 1 100 mL 60 g/ond n= 8. 33 x 163 redes - 0.0083 moles So, moles of coa if NaHCO₃ is limiting =10.003 modes – Actuel CH.coon = 0.0083 Holes. Limiting - Baking soda Trial B Nalco, 0.75 g n = 0.75g = 0.009 endes sugled CH₃ ooh w=8 mit n = sax 18 ml x 1 = 6.67 x 103 moles 60 g/nod = 0.0067 moles So, indes of co2 = 0.009 redes (Natio, limiting) = 0.0067 ndes) (Cloost lieting) Actual coll 100 ML [ Limiting Vinegar

1:59 Trial c NaHCO₃ n= 1.59 = 0.018 moles 84 god CH₂ cool v=5ml no. of moles - 59 x 5 ml x 1 = 4.167 x 10 redes 100 mL 60 g and -0.0042 endles & do, noles of coa = 0.018 modes (Natoz limiting) = 10:0042 moles (ch, coon finiting) Limiting Actual Vinegar í Bag in trial B will inflate the most with 0.0067 moles of co, All you have done is correct. Trial B Theoretical field = 0.0067 Co₂ Actuel yield = 0.0052 roles col % gield - 0.0052 x 100 =177.6% Trial C Theoretical field = 0.0042 ondes ₂ Actual yield = 0.0035 ender 7 yield - 0.0035 x 100 7 8) moles 9 O'0042 - [ 83.33%