Mass data for reaction of NaHCO3-Na2CO3 mixture with HCL Mass of crucible, cover

Question

Mass data for reaction of NaHCO3-Na2CO3 mixture with HCL

Mass of crucible, cover and residue after reaction with HCL (g)

1st weighing...

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

mass of NaCl formed (g)

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(%)

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

Explanation / Answer

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%

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---