The reform reaction between steam and gaseous methane () 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 liters per second of methane are consumed when the reaction is run at and . Calculate the r

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.

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.

Equation of reaction:

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,

That is 60.69 moles of hydrogen is produced per second.

Number of moles = Mass/ Molar mass

Rate at which H₂ is produced = 122.35 g/s = 0.12 kg/s

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

i believe it is a.) released into

explanation: