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2017-09-13 9:48 am
A 26.9 g sample of impure potassium nitrate (KNO3) was heated to complete decomposition according to the equation 2 KNO3(s) → 2 KNO2(s) + O2(g). After the reaction was complete, the solid residue (consisting of KNO2 and the original impurities) had a mass of 24.4 g. Assuming that only the potassium nitrate had decomposed, what was the percent KNO3 in the original sample?
回答 (2)
2017-09-13 10:06 am
Mass of O₂ formed = (26.9 - 24.4) g = 2.5 g
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
No. of moles of O₂ formed = (2.5 g) / (32.0 g/mol) = 0.0781 mol
2 KNO₃(s) → 2 KNO₂(s) + O₂(g)
Mole ratio KNO₃ : O₂ = 2 : 1
No. of moles of KNO₃ reacted = (0.0781 mol) × 2 = 0.156 mol
Molar mass of KNO₃ = (39.0 + 14.0 + 16.0×3) g/mol = 101.0 g/mol
Mass of KNO₃ reacted = (0.156 mol) × (101.0 g/mol) = 15.8 g
Percent KNO₃ in the original sample = (15.8/26.9) × 100% = 58.7%
OR:
Mass of KNO₃ in the sample = (26.9 - 24.4 g O₂) × (1 mol O₂ / 32.0 g O₂) × (2 mol KNO₃ / 1 mol O₂) × (101.0 g KNO₃ / 1 mol KNO₃) = 15.8 g KNO₃
Percent KNO₃ in the original sample = (15.8/26.9) × 100% = 58.7%
Molar mass of O₂ = 16.0 × 2 g/mol = 32.0 g/mol
No. of moles of O₂ formed = (2.5 g) / (32.0 g/mol) = 0.0781 mol
2 KNO₃(s) → 2 KNO₂(s) + O₂(g)
Mole ratio KNO₃ : O₂ = 2 : 1
No. of moles of KNO₃ reacted = (0.0781 mol) × 2 = 0.156 mol
Molar mass of KNO₃ = (39.0 + 14.0 + 16.0×3) g/mol = 101.0 g/mol
Mass of KNO₃ reacted = (0.156 mol) × (101.0 g/mol) = 15.8 g
Percent KNO₃ in the original sample = (15.8/26.9) × 100% = 58.7%
OR:
Mass of KNO₃ in the sample = (26.9 - 24.4 g O₂) × (1 mol O₂ / 32.0 g O₂) × (2 mol KNO₃ / 1 mol O₂) × (101.0 g KNO₃ / 1 mol KNO₃) = 15.8 g KNO₃
Percent KNO₃ in the original sample = (15.8/26.9) × 100% = 58.7%
2017-09-13 10:04 am
The difference in masses is due entirely to the loss of oxygen:
(26.9 g - 24.4 g) / (31.99886 g O2/mol) x (2 mol KNO3 / 1 O2) x (101.10332 g KNO3/mol) = 15.798 g KNO3
(15.798 g KNO3) / (26.9 g total) = 0.587 = 58.7% KNO3 by mass
(26.9 g - 24.4 g) / (31.99886 g O2/mol) x (2 mol KNO3 / 1 O2) x (101.10332 g KNO3/mol) = 15.798 g KNO3
(15.798 g KNO3) / (26.9 g total) = 0.587 = 58.7% KNO3 by mass
收錄日期: 2021-04-18 17:48:51
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