Question

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

Answer 1

Ans. Step 1: Let the mass of Na₂CO₃ be X grams and that of NaHCO₃ be Y grams in the given sample.

So,

            X gram + Y gram = 2.4822 gram

            Or, X + Y = 2.4822                         - equation 1

Now,

            Moles of Na₂CO₃ = Mass / molar mass

= (X g / 105.988736 g mol^-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 --------> 2 NaNO₃ + CO₂ + H₂O

            II. HNO₃ + NaHCO₃ --------> NaNO₃ + H₂O + CO₂

According to the stoichiometry of balanced reactions, 1 mol of Na₂CO₃ and 1 mol NaHCO₃ produces 1 mol of CO₂ each.

So,

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

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

# Step 3: Given, Volume of CO₂ produced = 0.644938 L

            Temperature = 294.25 K

            Pressure = 0.97012 atm

# Ideal gas equation:           PV = nRT     

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-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 L mol^-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 L mol^-1)

            Hence, n = 0.025899 mol

Therefore, total moles of CO₂ produced during the reaction = 0.025899 mol

Now, comparing #Step 2 and # Step 3, the total moles of CO₂ 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 = 1.0043 g

# Putting the value of Y in equation 1-

            X = 2.4822 – Y = 2.4822 – 1.00434 = 1.4779

Therefore, mass of Na₂CO₃ in sample = X g = 1.4779 g

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

                                                = (1.4779 g / 2.8422 g) x 100

                                                = 51.998 %

                                                = 52.0 %