Question

One way to make hydrogen for a fuel cell is by reforming hydrocarbon fuel. For example, propane can be turned into hydrogen in two steps. In the first step, propane (C3Hs) and water react to form carbon monoxide and hydrogen: C,H(+3H,01g)- 3cOe) + 7H,e In the second step, carbon monoxide and water react to form hydrogen and carbon dioxide: CO(g)+ H2O(g) H2(8)+CO2(8) Suppose the yield of the first step is 92.96 and the yield of the second step is 83, %. Calculate the mass of propane required to make 1.0 kg of hydrogen Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits

Answer 1

Let us try the reaction reverse

Balanced equation:
CO + H₂O =====> H₂ + CO₂
Reaction type: double replacement

Mass of Hydrogen = 1000 gm

Moles of Hydrogen = 1000 / 2.016 =  496.06 Moles

Yield of the reaction = 83 %

Moles of Hydrogen Theoritically produced = 496.06 x100 / 83 = 597.66 Moles

Moles of CO Theoritically produced = 597.66 Moles

First step reaction

Moles of CO actually produced = 597.66 Moles

Moles of propoane actually produced = 199.220 Moles

Yield of the reaction = 92 %

Moles of Propoane Theoritically produced = 216.544 Moles

Mass of Propane required = 216.544 x 44.1 =   9549.609 gm