Question

3.0012g of an unknown mixture of sodium carbonate (Na2CO3) and sodium bicarbonate (HNaCO3) is mixed with nitric acid (HNO3). 0.487277837 liters of CO2 is produced. What is the mass in grams and percent of carbonate (CO3^-) in the original sample (295.25K, 0.97012 atm). 486.18g of H2O was produced. Density of water at that temperature is 0.997747g/mL.

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 = 3.0012 gram

            Or, X + Y = 3.0012                         - 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 = 0.487277837 L

            Temperature = 295.25 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 0.487277837 L = n x (0.0821 atm L mol^-1K^-1) x 295.25 K

            Or, 0.47271797523044 atm L = n x 24.240025 atm L mol^-1

            Or, n = 0.47271797523044 atm L / (24.240025 atm L mol^-1)

            Hence, n = 0.0195015 mol

Therefore, total moles of CO₂ produced during the reaction = 0.0195015 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 = 0.0195015 mol

Or, 0.009434965X + 0.011903783Y = 0.0195015                   - equation 2

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

            0.009434965 X + 0.009434965Y = 0.028316216958

      (-) 0.009434965X + 0.011903783Y = 0.0195015

                                    -0.002468818 Y = + 0.008814716958

                                    Or, Y = 0.008814716958 / (-0.002468818)

                                    Hence, Y = -3.57042

Therefore, mass of NaHCO₃ in sample = Y g = (-)3.57042 g

# Note: Mass can’t be negative. So, there seems to be some discrepancies in the data mentioned in the question.

Error 2: The 3.0012 g sample of Na₂CO₃ + NaHCO₃ can never produce 468.18 g water. Note that moles of CO₂ produced is equal to moles of H₂O produced, as shown in step 2, balanced reaction.

So, maximum moles of H2O produced = 0.0195015 mol

Maximum mass of H2O produced = 0.0195015 mol x (18.0 g/ mol)

                                                            = 0.351027 g

Therefore, the volume of CO₂ and mass of H₂O produced are incorrectly mentioned in the question. Please get the correct values in question and follow the steps to calculate the mass of Na₂CO₃, and in turn moles and mass of CO₃^2- in the sample.