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1 February, 19:16

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.

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  1. 1 February, 21:32
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    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 ...
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