6. Bio-LPG is a co-product in the manufacture of bio-diesel from renewable crops and domestic and industrial waste - Leaving Cert Chemistry - Question 6 - 2021
Question 6
6. Bio-LPG is a co-product in the manufacture of bio-diesel from renewable crops and domestic and industrial waste. Hydrogen gas is a reagent in the production of th... show full transcript
Worked Solution & Example Answer:6. Bio-LPG is a co-product in the manufacture of bio-diesel from renewable crops and domestic and industrial waste - Leaving Cert Chemistry - Question 6 - 2021
Step 1
Give two commercial sources of hydrogen gas.
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Steam reforming of natural gas (coal gasification).
Electrolysis of water.
Step 2
State two main advantages of using hydrogen gas itself as a fuel.
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Hydrogen is low polluting, non-toxic, and produces no greenhouse gases during combustion.
It is efficient with a high energy density, providing high calorific value.
Step 3
What are the two main hydrocarbon components of LPG?
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Propane (C3H8)
Butane (C4H10)
Step 4
With the aid of a labelled diagram explain how crude oil is fractionally distilled.
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Crude oil is heated in a distillation column where fractions of different boiling points separate. Lighter fractions rise to the top and heavier fractions condense lower in the column. A labelled diagram will show the column with crude oil input at the bottom, and different fractions being drawn off at various heights, showing their respective boiling points.
Step 5
Show on your diagram where the refinery gas and gas oil fractions separate.
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On the labelled diagram, the refinery gas will condense at the upper part of the distillation column, while gas oil will be drawn from a lower point in the column.
Step 6
What is the purpose of adding mercaptans to natural gas or LPG?
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The purpose of adding mercaptans is to provide a distinctive smell to the gas, enabling leaks to be detected easily.
Step 7
Calculate the heat of formation of ethanol.
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Using the reaction:
ightarrow 2 ext{CO}_2(g) + 3 ext{H}_2 ext{O}(l) + ext{SO}_2(g)$$
The heat of formation of ethanol can be calculated as:
$$\Delta H_{f} = \Sigma \Delta H_{formation} – \Delta H_{combustion}$$
Substituting the values:
$$ \Delta H_{f} = (-787.0 - 857.4 - 296.8) - (-1867.5) \Rightarrow -1941.2 + 1867.5 = -73.7 \, kJ \, mol^{-1}$$
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