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Describe the Solvay process and related chemical reactions, including calculations for the reaction of sodium carbonate with sulfuric acid.

Chemistry

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  1. a)
    • i) Sodium chloride is obtained from sea water through solar evaporation. Sea water is channeled into large, shallow ponds (salt pans). The sun's heat causes the water to evaporate, leaving behind concentrated brine. As evaporation continues, sodium chloride crystals precipitate out and are then collected.
    • ii)
      • I. Equation for the formation of ammonium hydrogen carbonate: NH3(g)+H2O(l)+CO2(g)→NH4HCO3(aq)\text{NH}_3(\text{g}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g}) \to \text{NH}_4\text{HCO}_3(\text{aq})
      • II. Equation for the second reaction (formation of sodium bicarbonate): NH4HCO3(aq)+NaCl(aq)→NaHCO3(s)+NH4Cl(aq)\text{NH}_4\text{HCO}_3(\text{aq}) + \text{NaCl}(\text{aq}) \to \text{NaHCO}_3(\text{s}) + \text{NH}_4\text{Cl}(\text{aq})
    • iii)
      • I. Process I is filtration. This separates the solid sodium hydrogen carbonate from the ammonium chloride solution.
      • II. Process II is heating or thermal decomposition. This converts sodium hydrogen carbonate into sodium carbonate, water, and carbon dioxide.
    • iv)
      • I. Calculate the number of moles of sodium carbonate that reacted. The reaction between sodium carbonate and sulphuric acid is: Na2CO3(aq)+H2SO4(aq)→Na2SO4(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{SO}_4(\text{aq}) \to \text{Na}_2\text{SO}_4(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

  1. Calculate moles of sulphuric acid.

            $$
        \text{Moles of H}_2\text{SO}_4 = \text{Molarity} \times \text{Volume}
        $$
        $$
        \text{Moles of H}_2\text{SO}_4 = 0.5 \text{ M} \times \frac{40.0}{1000} \text{ L}
        $$
        $$
        \text{Moles of H}_2\text{SO}_4 = 0.02 \text{ mol}
        $$

  2. Use the mole ratio from the balanced equation.

            From the equation, 1 mole of $\text{Na}_2\text{CO}_3$ reacts with 1 mole of $\text{H}_2\text{SO}_4$.
        $$
        \text{Moles of Na}_2\text{CO}_3 = \text{Moles of H}_2\text{SO}_4
        $$
        $$
        \text{Moles of Na}_2\text{CO}_3 = 0.02 \text{ mol}
        $$
        The number of moles of sodium carbonate that reacted is $\boxed{\text{0.02 mol}}$.

    *   II. Determine the percentage of sodium carbonate in the sample.

  3. Calculate the molar mass of Na2CO3\text{Na}_2\text{CO}_3.

            $$
        \text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0)
        $$
        $$
        \text{Molar mass of Na}_2\text{CO}_3 = 46.0 + 12.0 + 48.0
        $$
        $$
        \text{Molar mass of Na}_2\text{CO}_3 = 106.0 \text{ g/mol}
        $$

  4. Calculate the mass of pure Na2CO3\text{Na}_2\text{CO}_3 in the sample.

            $$
        \text{Mass of Na}_2\text{CO}_3 = \text{Moles} \times \text{Molar mass}
        $$
        $$
        \text{Mass of Na}_2\text{CO}_3 = 0.02 \text{ mol} \times 106.0 \text{ g/mol}
        $$
        $$
        \text{Mass of Na}_2\text{CO}_3 = 2.12 \text{ g}
        $$

  5. Calculate the percentage purity.

            $$
        \text{Percentage purity} = \frac{\text{Mass of pure Na}_2\text{CO}_3}{\text{Mass of sample}} \times 100\%
        $$
        $$
        \text{Percentage purity} = \frac{2.12 \text{ g}}{2.15 \text{ g}} \times 100\%
        $$
        $$
        \text{Percentage purity} = 98.6046...\%
        $$
        The percentage of sodium carbonate in the sample is $\boxed{\text{98.60\%}}$.
    • b) Two industrial uses of sodium carbonate:
      • Manufacture of glass.
      • Manufacture of detergents and soaps.
      • Water softening (removing hardness from water).
    1. a) State the Charles' Law.
      • Charles' Law states that for a fixed mass of gas at constant pressure, the volume of the gas is directly proportional to its absolute temperature.
    • b) Calculate the temperature at which the volume of the gas would be 2.8×10−2 m32.8 \times 10^{-2} \text{ m}^3 at 1.0×105 Pascals1.0 \times 10^5 \text{ Pascals}. Given: V1=3.5×10−2 m3V_1 = 3.5 \times 10^{-2} \text{ m}^3 T1=291 KT_1 = 291 \text{ K} P1=1.0×105 PaP_1 = 1.0 \times 10^5 \text{ Pa} V2=2.8×10−2 m3V_2 = 2.8 \times 10^{-2} \text{ m}^3 P2=1.0×105 PaP_2 = 1.0 \times 10^5 \text{ Pa} We use the combined gas law: P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} Since P1=P2P_1 = P_2, the equation simplifies to Charles' Law: V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}

  6. Substitute the given values.

    $$
\frac{3.5 \times 10^{-2} \text{ m}^3}{291 \text{ K}} = \frac{2.8 \times 10^{-2} \text{ m}^3}{T_2}
$$

  7. Solve for T2T_2.

    $$
T_2 = \frac{(2.8 \times 10^{-2} \text{ m}^3) \times 291 \text{ K}}{3.5 \times 10^{-2} \text{ m}^3}
$$
$$
T_2 = \frac{2.8 \times 291}{3.5} \text{ K}
$$
$$
T_2 = 0.8 \times 291 \text{ K}
$$
$$
T_2 = 232.8 \text{ K}
$$
The temperature at which the volume would be $2.8 \times 10^{-2} \text{ m}^3$ is $\boxed{\text{232.8 K}}$.

    3. a) State the Graham's law diffusion. * Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (or density), at constant temperature and pressure.

    • b) The molar masses of gases W and X are 16.0 and 44.0 respectively. If the rate of diffusion of W through a porous material is 12cm3s−112 \text{cm}^3\text{s}^{-1}, calculate the rate of diffusion of X through the same material. According to Graham's Law of Diffusion: RateWRateX=MXMW\frac{\text{Rate}_W}{\text{Rate}_X} = \sqrt{\frac{M_X}{M_W}} Given: RateW=12 cm3s−1\text{Rate}_W = 12 \text{ cm}^3\text{s}^{-1} MW=16.0M_W = 16.0 MX=44.0M_X = 44.0

  8. Substitute the given values into the formula.

    $$
\frac{12 \text{ cm}^3\text{s}^{-1}}{\text{Rate}_X} = \sqrt{\frac{44.0}{16.0}}
$$
$$
\frac{12}{\text{Rate}_X} = \sqrt{2.75}
$$
$$
\frac{12}{\text{Rate}_X} \approx 1.6583
$$

  9. Solve for RateX\text{Rate}_X.

    $$
\text{Rate}_X = \frac{12}{1.6583}
$$
$$
\text{Rate}_X \approx 7.236 \text{ cm}^3\text{s}^{-1}
$$
The rate of diffusion of X is approximately $\boxed{\text{7.24 cm}^3\text{s}^{-1}}$.
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