chem help!?
2018-03-31 12:13 am
When 2.3 L of propane gas is completely combusted to form water vapor and carbon dioxide at 350°C and 0.902 atm, what mass of water vapor results?
回答 (1)
2018-03-31 2:03 am
Consider the C₃H₈ gas :
Pressure, P = 0.902 atm
Volume, V = 2.3 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 350) K = 623 K
Gas law: PV = nRT
No. of moles of C₃H₈ gas, n = PV/(RT) = 0.902 × 2.3 / (0.08206 × 623) mol = 0.04058 mol
Balanced equation for combustion of C₃H₈ gas :
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : H₂O = 1 : 4
No. of moles of H₂O resulted = (0.04058 mol) × 4 = 0.1623 mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Mass of H₂O resulted = (0.1623 mol) × (18.0 g/mol) = 2.92 g
Pressure, P = 0.902 atm
Volume, V = 2.3 L
Gas constant, R = 0.08206 L atm / (mol K)
Temperature, T = (273 + 350) K = 623 K
Gas law: PV = nRT
No. of moles of C₃H₈ gas, n = PV/(RT) = 0.902 × 2.3 / (0.08206 × 623) mol = 0.04058 mol
Balanced equation for combustion of C₃H₈ gas :
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Mole ratio C₃H₈ : H₂O = 1 : 4
No. of moles of H₂O resulted = (0.04058 mol) × 4 = 0.1623 mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Mass of H₂O resulted = (0.1623 mol) × (18.0 g/mol) = 2.92 g
收錄日期: 2021-04-24 00:59:33
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