Question

Table A: Mass data for reaction of NaHCO3 - Na2CO3 mixture with HCl:

Mass of crucible and cover	33.2868 (g)
Mass of crucible, cover, and mixture	33.7864 (g)
Mass of crucible, cover, and residue after reaction with HCl (1st weighing)	33.7070 (g)
Mass of crucible, cover, and residue after reaction with HCl (2nd weighing)	33.7012 (g)
Mass of crucible, cover, and residue after reaction with HCl (3rd weighing)	33.7020 (g)
Mass of unknown mixture used	0.4996 (g)

Using the data from Table A (above), solve the calculations in Table B (yellow) and answer questions (A-D). (Please show all steps in calculations so I can understand how to solve).

Table B: Calculation of percent of NaHCO₃ and Na₂CO₃ in an unknown mixture

Calculation of percent NaHCO3 and Na2CO3 in an unkown mixture (g)
Mass of unknown mixture used (g)
Mass of NaCl fomed (g)
Mass of NaHCO3 in unknown mixture (g)
Mass of Na2CO3 in unknown mixture (g)
Percent of NaHCO3 by mass in unknown mixture (%)
Percent of Na2CO3 by mass in unknown mixture (%)
Average percent by mass of other elements in unknown sample (%)

A) Write the two mathematical equations using the experimental quantities of your experiment (remember that both equations must have identical units for both sides of the two equations):

B) Solve the two equations and calculate the mass of NaHCO₃ and the mass of Na₂CO₃:

C) Calculate the percent NaHCO₃ in the mixture:

D) Explain how the calculated percent NaHCO₃ and Na₂CO₃ by mass would be affected if the initial mixture were wet with water. Indicate clearly your reason.

Answer 1

Reactions:

Mass of unknown mixture used = 33.7864 - 33.2868 = 0.4996 g

Average of residue mass + crucible + cover = (33.7070+33.7012+33.7020)/3 = 33.7034 g

Mass of NaCl formed = 33.7034 - 33.2868 = 0.4166 g

Moles of NaCl formed = 0.4166 / 58.5 = 0.00712

Moles of HCl used = 0.00712

Moles of Na⁺ = 0.00712

Moles of Na⁺ from NaHCO₃ = (1/3) x 0.00712 = 0.002373

Moles of NaHCO₃ = 0.002373

Mass of NaHCO₃ = 0.002373 x 84

Mass of NaHCO₃ = 0.199332 g

Mass of Na₂CO₃ = 0.4996 - 0.199332 = 0.3 g

Percentage of NaHCO₃ by mass in unknown mixture = (0.199332 / 0.4996) x 100 = 39.89 %

Percentage of Na₂CO₃ by mass in unknown mixture = 100-39.89 = 60.11 %