Question

Data, Calculations, and Results Part A: Reaction Stoichiometry Trial 2 Average Trial 1 Mass of NaHCO Mass of beaker +acetic acid + NaHCO3 + watch glass before reaction Mass reaction mixture + watch glass after reaction Mass of CO2 gasO 534 of beaker+ COr gas 0 534 a released Moles of CO2 gas Moles of CO2, corrected releasedla ooalo ter o.oa5 Moles of NaHCO, used 0.034 for amount dissolved (see below) moles NOTE: 'Calculate by adding 0.0040 moles to the "Moles of CO2 gas released". This correction accounts for the amount of CO2 that dissolves in 50 mL of aqueous solution. Calculate (and show your work for at least trial #1) for the following: 1) The number of moles of CO2 for Trial #1, 2, and 3. 2) The number of moles of CO2, corrected, for Trial #1, 2, and 3. 3) The number of moles of NaHCO3 for Trial #1, 2, and 3. Page 1 of 3

Answer 1

Equation Required is Na H C O_3 + C H_3 C O O H rightarrow Na O O C C H_3 + H_2 O + C O_2 (g) uparrow Trial # 1 Mass of C O_2 gas released = (Mass of beaker + acetic acid + Na H C O_3 + watch glass before reaction) - (Mass of beaker + reaction Mix + watch glass after reaction) \r\n = (223.002 - 222.468) g 0.534 g Moles of C O_2 gas released = Mass of C O_2 (g)/Molecular w t of C O_2 = 0.534/44 = 0.0121 Moles of C O_2 gas, connected amount = 0.0121 + 0.004 = 0.0161 moles \r\n Molecular weight of Na H C O_3 = (23 + 1 + 12 + (16 times 3)) g/mol 84 g/mol Moles of Na H C O_3 used = Mass of Na H C O_3/Mol middot w t of Na H C O_3