The density of liquid hydrogen is 71 g/L. The propellant tank for liquid hydroge

Question

The density of liquid hydrogen is 71 g/L. The propellant tank for liquid hydrogen has a capacity of 383,000 gallons (1 gallon 3.8 L) 02 The density of liquid oxygen is 1.14 kg/L. The propellant tank for liquid oxygen has a capacity of 143,000 gallons (1 gallon 3.8 L) 2 3 What is the molar ratio of hydrogen-to-oxygen in the propellant tanks? (moles H2 / moles 02) go ahead and calculate it again Which is the limiting reactant in this case? How many gallons of water would be produced when the reaction of the fuels stored in the propellant tanks goes to completion? (hint: density of liquid water is 1 g/mL) The value of AEp is per mole of O2 consumed. How much energy is produced when the reaction of the fuels stored in the propellant tanks goes to completion? Express your answer in kilojoules.

Explanation / Answer

To calculate the molar ratio of H2 / O2, first calculate the present moles of each component in each tank, with successive conversion relations:

moles H2: 383000 gal * (3.8 L / 1 gal) * (71 gr / 1 L) * (1 Kg / 1000 gr) * (1Kmol / 2 Kg) = 51666.7 Kmol H2 moles O2: 143000 gal * (3.8 L / 1 gal) * (1.14 Kg / L) * (1Kmol / 32Kg) = 1935.9 Kmol O2 H2 / O2 = 51666.7 / 1935.9 = 26.7 Kmol H2 / KmolO2 For the reaction: H2 + 1/2 O2 = H2O; what is the limit reagent, testing O2: 1935.9 Kmol O2 * (1mol H2 / 1/2 mol O2) = 3871.8 mol H2 required -------- O2 is reactive limit. Calculation of H2O production, with support in limit reagent: 1935.9 Kmol O2 * (1 mol H2O / 1/2 mol O2) = 3871.8 mol H2O produced 3871.8 mol H2O * (18 Kg / 1 Kmol) * (1L / 1 Kg) * (1 gal / 3.8 L) = 18340.1 gallons produced. Calculation of H2O formation energy: ?Hf = -241.9 Kj / mol * 3871.8 mol H2O = -936588.42 Kj.

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