# Question #bba7c

May 17, 2015

You'd need ${\text{1 m}}^{3}$ of $C {O}_{2}$ and ${\text{4 m}}^{3}$ of ${H}_{2}$ to produce that much methane.

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

$C {O}_{2 \left(g\right)} + \textcolor{red}{4} {H}_{2 \left(g\right)} \to C {H}_{4 \left(g\right)} + 2 {H}_{2} {O}_{\left(g\right)}$

Notice that 1 mole of carbon dioxide reacts with $\textcolor{red}{4}$ 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 $C {O}_{2}$ and of ${H}_{2}$ reacted

$1 \cancel{{\text{m"^3CH_4) * (color(red)(4)"m"^3H_2)/(1cancel("m"^3CH_4)) = color(green)("4 m}}^{3} {H}_{2}}$

and

$1 \cancel{{\text{m"^3CH_4) * ("1 m"^3CO_2)/(1cancel("m"^3CH_4)) = color(green)("1 m}}^{3} C {O}_{2}}$