Answer:
The rate at which dihydrogen gas is being produced = 0.018 kg/s
Explanation:
Firstly, we write the balanced equation for the production of the synthesis gas
CH₄ + H₂O → CO + 3H₂
The rate of consumption of CH₄ is 159 litres per second. With the reaction ran at T = 294°C and a pressure of 0.86 atm
Using the ideal gas equation, we can convert the volumetric rate of consumption of methane to molar rate of consumption
PV = nRT
PV' = n'RT
P = pressure = 0.86 atm = 87,139.5 Pa
V' = 159 L/s = 0.159 m³/s
n' = ?
R = molar gas constant = 8.314 J/mol.K
T = absolute temperature in Kelvin = 294°C = 567.15 K
87,139.5 × 0.159 = n' × 8.314 × 567.15
n' = (87,139.5 × 0.159) ÷ (8.314 × 567.15)
n' = 2.9383547773 mol/s = 2.938 mol/s
From the stoichiometry of this reaction,
1 mole of methane gives 3 moles of dihydrogen gas
2.938 mol/s of methane will correspond to (3 × 2.938) mol/s of dihydrogen gas, that is, 8.815 mol/s.
Mass flowrate = (molar flowrate) × (molar mass)
Molar flowrate = 8.815 mol/s
Molar mass of dihydrogen gas = 2 g/mol = 0.002 kg/mol
Mass flowrate = 8.815 × 0.002 = 0.0176301287 kg/s = 0.018 kg/s to 2 s.f.
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