2.4822g of an unknown mixture

2.4822g of an unknown mixture of sodium carbonate (Na2CO3) andsodium bicarbonate (HNaCO3) is mixed with nitric acid (HNO3).0.644938 liters of CO2 is produced. What is the mass and percent ofsodium carbonate in the original sample (294.25K, 0.97012 atm).


Ans. Step 1: Let the mass ofNa2CO3 be X grams and that ofNaHCO3 be Y grams in the given sample.


           X gram + Y gram = 2.4822 gram

           Or, X + Y = 2.4822                       – equation 1


           Moles of Na2CO3 = Mass / molar mass

= (X g / 105.988736 gmol-1)

                                   = (X / 105.988736) mol

                                   = 0.009434965X mol

Moles of NaHCO3 = (Y / 84.006908) mol =0.011903783Y mol

# Step 2: Balanced reactions:

           I. Na2CO3 + 2HNO3 ——–> 2NaNO3 + CO2 + H2O

           II. HNO3 + NaHCO3 ——–>NaNO3 + H2O + CO2

According to the stoichiometry of balanced reactions, 1 mol ofNa2CO3 and 1 mol NaHCO3 produces 1mol of CO2 each.


Moles of CO2 produced by Na2CO3= 1 x moles of Na2CO3 = 0.009434965X mol

Moles of CO2 produced by NaHCO3 = 1 xmoles of NaHCO3 = 0.011903783Y mol Total moles ofCO2 produced = 0.009434965X mol + 0.011903783Ymol

# Step 3: Given, Volume of CO2produced = 0.644938 L

           Temperature = 294.25 K

           Pressure = 0.97012 atm

# Ideal gas equation:          PV =nRT     

           Where, P = pressure in atm

           V = volume inL                   

           n = number of moles

           R = universal gas constant= 0.0821 atm Lmol-1K-1

           T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in above equation-

0.97012 atm x 0.644938 L = n x (0.0821 atm Lmol-1K-1) x 294.25 K

           Or, 0.62566725256 atm L = n x 24.157925 atm L mol-1

           Or, n = 0.62566725256 atm L / (24.157925 atm Lmol-1)

           Hence, n = 0.025899 mol

Therefore, total moles of CO2 produced during thereaction = 0.025899 mol

Now, comparing #Step 2 and # Step 3, the total moles ofCO2 produced during reaction is given by –

0.009434965X mol + 0.011903783Y mol = 0.025899 mol

Or, 0.009434965X + 0.011903783Y =0.025899                   – equation 2

# Step 4: Comparing (equation 1 x 0.009434965)– equation 2-

           0.009434965X + 0.009434965Y = 0.02341947

      (-) 0.009434965X+ 0.011903783Y = 0.025899

                                   -0.002468818 Y = -0.002479530

                                   Or, Y = 0.002479530 / 0.002468818

                                   Hence, Y = 1.0043

Therefore, mass of NaHCO3 in sample= Y g = 1.0043g

# Putting the value of Y in equation 1-

           X = 2.4822 – Y = 2.4822 – 1.00434 = 1.4779

Therefore, mass of Na2CO3 insample = X g = 1.4779 g

# Step 5: % (w/w) Na2CO3in sample = (Mass of Na2CO3 / Mass of sample)x 100

                                               = (1.4779 g / 2.8422 g) x 100

                                               = 51.998 %

                                               = 52.0 %

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