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Question 19M.3.sl.TZ1.13b

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Date May 2019 Marks available [Maximum mark: 3] Reference code 19M.3.sl.TZ1.13b
Level sl Paper 3 Time zone TZ1
Command term Show Question number b Adapted from N/A
b.
[Maximum mark: 3]
19M.3.sl.TZ1.13b

E10 is composed of 10 % ethanol and 90 % normal unleaded fuel.

(b)

Show that, for combustion of equal masses of fuel, ethanol (Mr = 46 g mol−1) has a lower carbon footprint than octane (Mr = 114 g mol−1).

[3]

Markscheme

Alternative 1
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) / 1 mol ethanol produces 2 mol CO2
OR
C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (l) / 1 mol octane produces 8 mol CO2  [✔]


For 1 g of fuel:
«  1g 46 g mo l − 1 × 2 mol CO2 (g) =» 0.04 «mol CO2 (g)» from ethanol  [✔]

«  1g 114 g mo l − 1 × 8 mol CO2 (g) =» 0.07 «mol CO2 (g)» from octane  [✔]

 

Alternative 2
ratio of C in ethanol:octane is 2:8, so ratio in carbon dioxide produced per mole will be 1:4  [✔]

ratio amount of fuel in 1 g =  1 46 : 1 114 = 2.5:1  [✔]

4 > 2.5 so octane produces more carbon dioxide
OR
ratio of amount of carbon dioxide = 2.5:4 = 1:1.61 so octane produces more «for combustion of same mass» [✔]

Examiners report

A question that gave the opportunity for a variety of different approaches. This challenge was beyond all but the best students, though there were a number of well argued responses.

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