Question

Question #bba7c

Answer 1

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)`