Question

Mass data for reaction of NaHCO3-Na2CO3 mixture with HCL

Mass of crucible and cover (g)	28.8258
Mass of crucible, cover and mixture (g)	29.4948
Mass of crucible, cover and residue after reaction with HCL (g) 1st weighing...	29.4454
2nd weighing...	29.4455

Calculation of percent NaHCO3 and Na2CO3 in an unknown mixture. ( I need help filling out this chart)

mass of unkown mixture used (g)
mass of NaCl formed (g)
Mass of NaHCO3 in unkown mixture (g)
Mass of Na2CO3 in unknown micture (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 (%)

Write 2 mathematical equations using the experiemental quanitities of your expereriment. You must remember that both equations must have identical units or both sides of the equation.

** please solve**

1. Mass of mixture = mass NaHCO3 + Mass Na2CO3

2. Moles NaCL= 2 (moles NaCO3)+ 1(mol NaHCO3)

3. mass NAHCO3= mass mixture- Na2CO3

given : MW Na2CO3= 105.98 g/mol MW NaCL= 58.44 g/mol MW NaHCO3=84.01 g/mol Mass of mixture =? Mass of NaCl=?

SOLVE the 2 equations and calculate the mass of NaHCO3 and the mass of Na2CO3. (showing work please!)

Calculae the percent NaHCO3 by mass in the unknown mixture, Calculate the percent Na2CO3 by mass in the unknown mixture, and the average percent by mass of other elements in unknown sample(%)

Answer 1

NaHCO₃ and Na₂CO₃ with HCl-

Mass of crucible and cover = 28.8258 gm

Mass of crucible, cover and mixture = 29.4948 gm

Mass of crucible, cover and residue after reaction with HCl (I Weighing) = 29.4454 gm

Mass of crucible, cover and residue after reaction with HCl (II Weighing) = 29.4455 gm

So, Mass of Mixture = (Mass of crucible, cover and mixture) - (Mass of crucible and cover)

= (29.4948 - 28.8258) gm = 0.669 gm

NaHCO₃ and Na₂CO₃ in an Unknown Mixture-

Mass of Unknown Mixture Used = gm

Mass of NaCl formed = gm

Mass of NaHCO₃ in Unknown Mixture = gm

Mass of Na₂CO₃ in Unknown Mixture = gm

Percent of NaHCO₃ in Unknown Mixture = %

Percent of Na₂CO₃ in Unknown Mixture = %

Average Percent by Mass of other elements in Unknown Sample = %

We have to find out -----

1.) Mass of Mixture = Mass of NaHCO₃ + Mass of Na₂CO₃

2.) Moles NaCl = 2 (moles NaCO₃ ) + 1 (mole NaHCO₃)

3.) Mass NaHCO₃ = Mass Mixture - Mass of Na₂CO₃

MW Na₂CO₃ = 105.98 g/mol

MW NaCl = 58.44 g/mol

MW NaHCO₃ = 84.01 g/mol

Since, molecular weight and molar mass are same, therefore,

2 HCl (aq) + Na₂CO₃ (aq) → 2 NaCl (aq) + NaHCO₃ (aq)

NAHCO₃ + HCl -----> NACl + CO₂ + H₂O

1.)  Mass of Mixture = Mass of NaHCO₃ + Mass of Na₂CO₃

   ₌ 84.01 g/mol + 105.98 g/mol = 189.99 g/mol

2.) Moles NaCl = 2 (moles Na₂CO₃ ) + 1 (mole NaHCO₃)

Moles Na₂CO₃ = mass Na₂CO₃/ molar mass Na₂CO₃

   = 1/ 105.98 = 0.0094

Moles NaHCO₃ = mass NaHCO₃/ molar mass NaHCO₃

= 1 / 84.01 = 0.0119

Moles NaCl = 2 (0.0094) + 0.0119

= 0.0188 + 0.0119 = 0.0307

3.) Mass NaHCO₃ = Mass Mixture - Mass of Na₂CO₃

   = 189.99 g/mol - 105.98 g/mol = 84.01 g/mol

Percent of Na₂CO₃ = (Mass/Total Mass) x 100%

Total Mass =  189.99 g/mol

Therefore, Percent by Mass of Na₂CO₃ = [Mass of Na₂CO₃ /Total Mass] x 100%

= [105.98 g/mol / 189.99 g/mol] x 100%

= 55.78%

Percent by Mass of NaHCO₃ = [84.01 g/mol / 189.99 g/mol] x 100%

= 44.218%