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Zorluk: Çok zorAlkaline Earth Metals: Calcium Extraction, Properties, and Compounds
A 25.0 g25.0\text{ g} sample of limestone containing 80.0%80.0\% calcium trioxocarbonate(IV) by mass is strongly heated until decomposition is complete according to the equation:
CaCO3(s)→ΔCaO(s)+CO2(g)CaCO_3(s) \xrightarrow{\Delta} CaO(s) + CO_2(g)
What volume of carbon(IV) oxide gas, measured at room temperature and pressure (RTP), is liberated in this reaction?
(M(CaCO3)=100 g mol−1M(CaCO_3) = 100\text{ g mol}^{-1}; Molar volume of gas at RTP =24.0 dm3 mol−1= 24.0\text{ dm}^3\text{ mol}^{-1})
  1. 4.80 dm34.80\text{ dm}^3Cevap
  2. B
    4.48 dm34.48\text{ dm}^3
  3. C
    6.00 dm36.00\text{ dm}^3
  4. D
    5.60 dm35.60\text{ dm}^3

Cevap

The volume of carbon(IV) oxide gas liberated at RTP is 4.80 dm34.80\text{ dm}^3.
The mass of active CaCO3CaCO_3 is 80.0%80.0\% of 25.0 g25.0\text{ g}, which equals 20.0 g20.0\text{ g}. Dividing by the molar mass (100 g mol−1100\text{ g mol}^{-1}) gives 0.20 mol0.20\text{ mol} of CaCO3CaCO_3. By stoichiometry, 0.20 mol0.20\text{ mol} of CO2CO_2 is evolved. Multiplying by the molar volume at RTP (24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}) yields 4.80 dm34.80\text{ dm}^3.

Adım Adım Çözüm

1
Calculate the mass of pure calcium trioxocarbonate(IV) (CaCO3CaCO_3) present in the limestone sample.
Mass of pure CaCO3=80.0100×25.0 g=20.0 gCaCO_3 = \frac{80.0}{100} \times 25.0\text{ g} = 20.0\text{ g}.
Impurities in the limestone do not produce CO2CO_2 gas upon heating.
2
Determine the amount in moles of pure CaCO3CaCO_3 decomposed.
Moles of CaCO3=20.0 g100 g mol−1=0.20 molCaCO_3 = \frac{20.0\text{ g}}{100\text{ g mol}^{-1}} = 0.20\text{ mol}.
Converting mass to amount in moles allows stoichiometric evaluation using balanced chemical equations.
3
Use the stoichiometric ratio from the balanced chemical equation to find the moles of CO2CO_2 produced.
Since 1 mol CaCO3→1 mol CO21\text{ mol } CaCO_3 \rightarrow 1\text{ mol } CO_2, moles of CO2=0.20 molCO_2 = 0.20\text{ mol}.
The thermal decomposition ratio between CaCO3CaCO_3 and CO2CO_2 is 1:11:1.
4
Calculate the volume of CO2CO_2 gas at room temperature and pressure (RTP).
Volume of CO2=0.20 mol×24.0 dm3 mol−1=4.80 dm3CO_2 = 0.20\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 4.80\text{ dm}^3.
Molar gas volume at RTP is 24.0 dm3 mol−124.0\text{ dm}^3\text{ mol}^{-1}.

Anahtar Kavram

Thermal decomposition of calcium carbonate and percentage purity stoichiometry at non-STP conditions
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Alkaline Earth Metals: Calcium Extraction, Properties, and Compounds

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