Question

2) Calculations to determine the amount of CO2(g) produced in the Baking Soda-Vinegar reaction
Starting Material: 40 ml Vinegar - SpecificGrav (CH3COOH) = 1.05 &  vol%(CH3COOH) = 5 vol%
a) Calculate the stoichiometric amount (moles and grams) of sodium bicarbonate needed.

[CH₃COOH(aq.) + H₂O(l)]+ NaHCO₃(s) --> H₂O(l) +CO₂(g) + NaCH₃COO(aq.)

c) Given that 1 mol of an ideal gas at standard temperature and pressure occupies 22.4 L,
calculate the Ideal Gas Constant (R) in units of liter-atm/mol-K.
NOTE: Remember R is calculated at Standard Temperature and Pressure (STP = 273 K, 1atm)

d) Use the Ideal Gas Law to calculate the volume of CO2 you expect to be produced from the
stoichiometric reaction of your 40 mL of vinegar at the room temperature (e.g. probably
roughly T = 20 `C). How many moles of CO2 is that?

e) What would be the pressure in a 700 mL Propel plastic sports drink bottle if you ran this
reaction inside it with the valve cap closed?

f) Limiting Reactant Concept
Given 75 mL of vinegar and 8 g baking soda (BS), which is the limiting reactant (i.e. the
limiting reactant is the reactant that would be fully consumed by the reaction while other
reactants are present in excess and would still have unreacted material present after the
reaction)? Show calculations. What would be the volume of CO2 gas produced and the
pressure in the Propel bottle under these conditions? Show calculations.

Answer 1

CLASSTIME Page No CH₂COOH + Naticoz cu, wona + H2O + CO₂ (g) 40 me 5.1.6 Vol ₃ COOH density of CH3COOH = 1.05 g met 100 me of solution contains 5 ml CH₃COOH 40 me of solution " 15 x 40 me alcool (oo Mass of CH₃COOH in 40 me = 2 me CH₃Cook solution = 2 me x 1.05 g met = 2.1 moles of 09 molt the reaction 1 mole of T = 0.035 mol tells us the one mole of Colla GDH produces co2 0.035 mol of co will be produced. (b) to completely react cuz wol & Nayco required = 0.05 mol of Na Hoz is. 0.035 molx 84 gmol 2949 G n=1 mol P= I atm v= 22.4L T= 273k PV = nRT R2 PV 1 atmx 224 L ИТ 1 mol X 23K PV - 008205 Latm k'mol

nat oom moles of temp co, & 298K (20°c) 0.035 mol = pe 1 atm » V = nRT 2 0.035 X 0.08205 X 293 XP 1 0.8914 L - 891.4 me co, produced. If reaction takes place in 500 me bottle P = MRT V 0.035 X 0.08905 x 293 w. 700 1000 core 1.202 atm. 0001 75 me vinegar reaboda mass of CH₂ WOH = 5 75 A I & 3.75 ml x 1.05 g met 100 = 3.9375 g in 75 me vinegar = 0.0656 mol moles mores of watco in sg 0.0952 mol - 84 quatt since 1 mole of thus cu Cool be consumed CH₂ WOH present first react with I mole Natco in small amount will and will determine the amount of product. I Limiting reagent Excees Reagent – CHg won. - NaHCO₃

CLASSTIME Page No 0.0656 mol CH₃COOH produce 0.0656 mol coa volume of co, (at ) = NRI 20con P = 0.0656X 0.0821 x 293 = 1.577 L : 1.577 LA Pressure bottle 20 MRI 0.0656 x 0.0821 x 293 700 1000 3 2.9529 atm . V