Question

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

Answer:

CH4 + H2O -> CO + 3H2

We use PV = nRT to find moles of CH4 for 1 second, T = 230C = 230 + 273 = 503 K

0.35 atm x 246 L = nx 0.08206 liters atm / molK x 503 K

n = 2,086

Moles of H2 produced = 3 x moles of CH4 = 3 x 2,086 = 6,258

Mass of H2 produced = moles x molar mass of H2 = 6,258 mol x 2 g / mol = 12.52 g = 0.0125 kg

Thus H2 products = 0.013 kg / s (rounded to two digits)

 

Explanation:

We decided to use the equation PV = nRT since the pressure, the temperature vary, but considering it to be a noble gas, that is why we take into account the variable R, which is a constant of the noble gases, its value is 0.082 with its respective units. .