Question #0c07e

1 Answer
Ernest Z.
Feb 7, 2016

The combustion consumes 1070 mL of #"O"_2#.

Explanation:

We don't even have to use moles or the Ideal Gas Law to solve this problem.

We can use Gay-Lussac's Law of Combining Volumes — when gases react to form other gases at the same temperature and pressure, they do so in whole-number ratios.

#"2C"_6"H"_14 + "19O"_2("g") → "12CO"_2("g") + "14H"_2"O"#
#color(white)(mmmmm)"19 mL"color(white)(lmml) "12 mL"#

The balanced equation tells us that 19 mL of #"O"_2# are required to form 12 mL of #"CO"_2#.

∴ #676 color(red)(cancel(color(black)("mL CO"_2))) × ("19 mL O"_2)/(12 color(red)(cancel(color(black)("mL CO"_2)))) = "1070 mL O"_2#

Related questions
See all questions in Gas Stoichiometry
Impact of this question
1835 views around the world
You can reuse this answer
Creative Commons License