Question #bba7c

1 Answer
Stefan V.
May 17, 2015

You'd need #"1 m"^3# of #CO_2# and #"4 m"^3# of #H_2# to produce that much methane.

Once again, start with the balanced chemical equation for the Sabatier reaction

#CO_(2(g)) + color(red)(4)H_(2(g)) -> CH_(4(g)) + 2H_2O_((g))#

Notice that 1 mole of carbon dioxide reacts with #color(red)(4)# moles of hydrogen to produce 1 mole of methane.

Because all the gases are under the same conditions for pressure and temperature, you can say that 1 cubic meter of carbon dioxide will react with 4 cubic meters of hydrogen to produce 1 cubic meter of methane.

Since you want your reaction to produce 1 cubic meter of methane, you can work backwards to determine how many liters of #CO_2# and of #H_2# reacted

#1cancel("m"^3CH_4) * (color(red)(4)"m"^3H_2)/(1cancel("m"^3CH_4)) = color(green)("4 m"^3H_2)#

and

#1cancel("m"^3CH_4) * ("1 m"^3CO_2)/(1cancel("m"^3CH_4)) = color(green)("1 m"^3CO_2)#

Related questions
See all questions in Gas Stoichiometry
Impact of this question
1790 views around the world
You can reuse this answer
Creative Commons License