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Zorluk: OrtaPercentage Composition and Percentage Purity Calculations
A 5.00 g5.00\text{ g} sample of impure sodium trioxocarbonate(IV), Na2CO3\text{Na}_2\text{CO}_3, reacted completely with excess dilute hydrochloric acid to produce 0.896 dm30.896\text{ dm}^3 of carbon(IV) oxide gas at STP according to the equation:
Na2CO3(s)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3(s) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)
What is the percentage purity of the sodium trioxocarbonate(IV) sample? [Molar volume of gas at STP=22.4 dm3mol−1,Na=23,C=12,O=16][\text{Molar volume of gas at STP} = 22.4\text{ dm}^3\text{mol}^{-1}, \text{Na} = 23, \text{C} = 12, \text{O} = 16]
  1. 84.8%Cevap
  2. B
    79.1%
  3. C
    66.4%
  4. D
    42.4%

Cevap

The percentage purity of the sodium trioxocarbonate(IV) sample is 84.8%.
The correct answer of 84.8% is obtained by converting the volume of carbon(IV) oxide gas at STP to moles using the molar volume of 22.4 dm³/mol, determining the equivalent mass of pure sodium trioxocarbonate(IV) using its molar mass (106 g/mol), and finding its proportion relative to the 5.00 g sample mass.

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1
Calculate the number of moles of carbon(IV) oxide gas produced at STP.
n(CO2)=0.896 dm322.4 dm3mol−1=0.04 moln(\text{CO}_2) = \frac{0.896\text{ dm}^3}{22.4\text{ dm}^3\text{mol}^{-1}} = 0.04\text{ mol}
At STP, one mole of any gas occupies a volume of 22.4 dm322.4\text{ dm}^3.
2
Determine the molar mass and the mass of pure sodium trioxocarbonate(IV).
Molar mass of Na2CO3=(2×23)+12+(3×16)=106 g/mol\text{Na}_2\text{CO}_3 = (2 \times 23) + 12 + (3 \times 16) = 106\text{ g/mol}. Mass of pure Na2CO3=0.04 mol×106 g/mol=4.24 g\text{Na}_2\text{CO}_3 = 0.04\text{ mol} \times 106\text{ g/mol} = 4.24\text{ g}.
From the balanced chemical equation, 1 mol1\text{ mol} of Na2CO3\text{Na}_2\text{CO}_3 yields 1 mol1\text{ mol} of CO2\text{CO}_2.
3
Calculate the percentage purity of the sample.
Percentage Purity=(Mass of pure Na2CO3Mass of impure sample)×100%=(4.24 g5.00 g)×100%=84.8%\text{Percentage Purity} = \left(\frac{\text{Mass of pure } \text{Na}_2\text{CO}_3}{\text{Mass of impure sample}}\right) \times 100\% = \left(\frac{4.24\text{ g}}{5.00\text{ g}}\right) \times 100\% = 84.8\%.
Percentage purity expresses the mass of the pure reacting component relative to the total sample mass.

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Percentage Purity from Gas Stoichiometry
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