2.4822g of an unknown mixture

Ans. Step 1: Let the mass ofNa₂CO₃ be X grams and that ofNaHCO₃ 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 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.

So,

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 %