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Difficulty: Very hardOxides of Carbon and Trioxocarbonate(IV) Salts

A solid mixture containing 16.8 g16.8\text{ g} of NaHCO3\text{NaHCO}_3 and 10.6 g10.6\text{ g} of Na2CO3\text{Na}_2\text{CO}_3 is heated strongly in an open crucible until no further change in mass occurs. Given the molar volume of any gas at s.t.p. is 22.4 dm3mol−122.4\text{ dm}^3\text{mol}^{-1} and relative atomic masses (Na=23,H=1,C=12,O=16)(\text{Na}=23, \text{H}=1, \text{C}=12, \text{O}=16), what is the volume of CO2\text{CO}_2 gas evolved at s.t.p. and the total mass of the solid residue remaining?

  1. 2.24 dm32.24\text{ dm}^3 of CO2\text{CO}_2 and 21.2 g21.2\text{ g} of solid residueAnswer
  2. B
    4.48 dm34.48\text{ dm}^3 of CO2\text{CO}_2 and 10.6 g10.6\text{ g} of solid residue
  3. C
    2.40 dm32.40\text{ dm}^3 of CO2\text{CO}_2 and 21.2 g21.2\text{ g} of solid residue
  4. D
    4.48 dm34.48\text{ dm}^3 of CO2\text{CO}_2 and 12.4 g12.4\text{ g} of solid residue

Answer

The volume of carbon(IV) oxide evolved at s.t.p. is 2.24 dm32.24\text{ dm}^3 and the total mass of the solid residue remaining is 21.2 g21.2\text{ g}.
Heating 16.8 g16.8\text{ g} (0.20 mol0.20\text{ mol}) of NaHCO3\text{NaHCO}_3 yields 0.10 mol0.10\text{ mol} of CO2\text{CO}_2 gas, which occupies 2.24 dm32.24\text{ dm}^3 at s.t.p. (0.10×22.4 dm30.10 \times 22.4\text{ dm}^3). The reaction produces 0.10 mol0.10\text{ mol} (10.6 g10.6\text{ g}) of solid Na2CO3\text{Na}_2\text{CO}_3. Adding this to the unreacted 10.6 g10.6\text{ g} of original Na2CO3\text{Na}_2\text{CO}_3 yields a total solid residue mass of 21.2 g21.2\text{ g}.

Step-by-Step Solution

1
Calculate the molar masses of the relevant substances
Molar mass of NaHCO3=23+1+12+(3×16)=84 g/mol\text{Molar mass of NaHCO}_3 = 23 + 1 + 12 + (3 \times 16) = 84\text{ g/mol}; Molar mass of Na2CO3=(2×23)+12+(3×16)=106 g/mol\text{Molar mass of Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}.
Molar masses are needed to convert mass to chemical amounts (moles).
2
Determine thermal stability of components and identify the decomposition reaction
Sodium trioxocarbonate(IV) (Na2CO3\text{Na}_2\text{CO}_3) is thermally stable and does not decompose. Sodium hydrogentrioxocarbonate(IV) decomposes: 2NaHCO3(s)→ΔNa2CO3(s)+H2O(g)+CO2(g)2\text{NaHCO}_3(s) \xrightarrow{\Delta} \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g).
Alkali metal trioxocarbonates(IV) (except lithium) do not decompose on heating, whereas hydrogentrioxocarbonates(IV) decompose to form trioxocarbonate(IV), water vapor, and carbon(IV) oxide.
3
Calculate the moles of NaHCO3\text{NaHCO}_3 and the volume of CO2\text{CO}_2 evolved
Moles of NaHCO3=16.8 g84 g/mol=0.20 mol\text{Moles of NaHCO}_3 = \frac{16.8\text{ g}}{84\text{ g/mol}} = 0.20\text{ mol}. From stoichiometry, 2 mol NaHCO3→1 mol CO22\text{ mol NaHCO}_3 \rightarrow 1\text{ mol CO}_2. Thus, moles of CO2=0.202=0.10 mol\text{moles of CO}_2 = \frac{0.20}{2} = 0.10\text{ mol}. Volume of CO2 at s.t.p.=0.10 mol×22.4 dm3mol−1=2.24 dm3\text{CO}_2 \text{ at s.t.p.} = 0.10\text{ mol} \times 22.4\text{ dm}^3\text{mol}^{-1} = 2.24\text{ dm}^3.
Stoichiometric mole ratio determines the yield of gaseous product at standard temperature and pressure.
4
Calculate the total mass of the solid residue remaining
From decomposition: moles of new Na2CO3=0.10 mol\text{moles of new Na}_2\text{CO}_3 = 0.10\text{ mol}. Mass of new Na2CO3=0.10 mol×106 g/mol=10.6 g\text{Na}_2\text{CO}_3 = 0.10\text{ mol} \times 106\text{ g/mol} = 10.6\text{ g}. Total solid residue = original Na2CO3\text{Na}_2\text{CO}_3 + produced Na2CO3=10.6 g+10.6 g=21.2 g\text{Na}_2\text{CO}_3 = 10.6\text{ g} + 10.6\text{ g} = 21.2\text{ g}.
The solid residue consists of both the thermally stable initial component and the newly formed trioxocarbonate(IV) salt.

Key Concept

Thermal decomposition of group 1 hydrogentrioxocarbonate(IV) salts vs trioxocarbonate(IV) salts and gas stoichiometry.
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Topics

Oxides of Carbon and Trioxocarbonate(IV) Salts

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