Given: 2.99g of NaHCO3 reacts with 30 mL of CH3COOH and is heated to produce 24.
ID: 858587 • Letter: G
Question
Given: 2.99g of NaHCO3 reacts with 30 mL of CH3COOH and is heated to produce 24.77g of NaC2H3O2
What is the balanced equation for the reaction of baking soda and acetic acid.
Using the balanced equation and the amount of NaHCO3, determine the theoretical yield of NaC2H3O2 in moles and grams. Find the percent yield for the reaction.
What is the theoretical yield for CO2? Using the density value, 1.25 g/L, calculate what volume of CO2 would be produced by this reaction in an oven. Show your calculations.
How many moles of NaHCO3 and HC2H3O2 are necessary to produce 425mL of CO2 at baking temperature? Show your calculations. Be sure to include the percent yield for this reaction in the calculations.
Explanation / Answer
1)
NaHCO3 + CH3COOH ----> CH3COONa + H2O + CO2
No of moles of NaHCO3 = 2.99/84.007 = 0.0356 moles
No of moles of CH3COOH = 17.04*30/1000 =
theoretically 1 mole NaHCO3 = 1mole CH3COONa
Theortical yield of CH3COONa = 0.0356 mole.
Theortical yield of CH3COONa mass = 0.0356 *82.03 = 2.92 grams
practical yield
No of moles of CH3COONa produced = 24.77/82.03 = 0.302 mole
percent yield = practical yield/Theoretical yield/ *100
% yield = 0.302 / 0.0356*100 = 848.314606742 %
2) Theortical yield of CO2 = 0.0356*44 = 1.5664 grams
volume of CO2 = M/D = 1.5664/1.25 = 1.25 L
3) Density of CO2 = 1.25 g/L
mass of Co2 = d*m = 0.425*1.25 = 0.53125 grams
no of moles of CO2 = 0.53125/44 = 0.012 mole
No of moles of NaHCO3 required = 0.012 mole
No of moles of HC2H3O2 required = 0.012 mole
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