Question

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 rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to significant digits..

Answer 1

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: The rate at which dihydrogen is being produced is 0.12 kg/sec

Step-by-step explanation:

The balanced chemical equation is ;

`CH_4+H_2O\rightarrow 3H_2+CO`

According to ideal gas equation:

`PV=nRT`

P = pressure of gas = 0.96 atm

V = Volume of gas = 924 L

n = number of moles

R = gas constant =
`0.0821Latm/Kmol`

T =temperature =
`261^0C=(261+273)K=534K`

`n=(PV)/(RT)`

`n=(0.96atm* 924L)/(0.0820 L atm/K mol* 534K)=20.2moles`

According to stoichiometry:

1 mole of methane produces = 3 moles of hydrogen

Thus 20.2 moles of methane produces =
`(3)/(1)* 20.2=60.6` moles of hydrogen

Mass of hydrogen =
`moles* {\text {Molar mass}}=60.6mol* 2g/mol=121.2g=0.12kg`

Thus the rate at which dihydrogen is being produced is 0.12 kg/sec