Respuesta :
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.
Hope this Helps!!!
Complete Question:
The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.
Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924. liters per second of methane are consumed when the reaction is run at 261.°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
Answer:
Rate at which H₂ is produced = 0.12 kg/s
Explanation:
Volume of methane produced, V = 924 litres
Temperature, T = 261.°C = 261 + 273
T = 534 K
Pressure, P = 0.96 atm
Gas constant, R = 0.0821 L-atm/ K-mol
We will calculate the number of moles of methane used in the reaction.
[tex]n_{methane} = \frac{PV}{RT} \\n_{methane} = \frac{0.96 * 924}{0.0821 * 534} \\n_{methane} = \frac{887.04}{43.8414}\\n_{methane} = 20.23 moles[/tex]
Equation of reaction:
[tex]CH_{4} + H_{2} O \rightarrow CO + 3H_{2}[/tex]
From the reaction above :
1 mole of methane produced 3 moles of H₂
20.23 moles of methane will produce (20.23 * 3 ) moles of H₂
Number of moles of H₂, [tex]n_{H_{2} } = 60.69 moles[/tex]
That is 60.69 moles of hydrogen is produced per second.
Number of moles = Mass/ Molar mass
[tex]n_{H_{2} } = \frac{Mass_{H_{2} }}{Molar mass_{H_{2} }} \\Mass_{H_{2}} = n_{H_{2} } * Molar mass_{H_{2} }\\Mass_{H_{2}} = 60.69 * 2.016\\Mass_{H_{2}} = 122.35 g[/tex]
Rate at which H₂ is produced = 122.35 g/s = 0.12 kg/s
