Question #bba7c

Question

Question #bba7c

1 Answer

Impact of this question

2019 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-17T09:18:06

You'd need ${1 m}^{3}$ $"1 m"^3$ of $C O_{2}$ $CO_2$ and ${4 m}^{3}$ $"4 m"^3$ of $H_{2}$ $H_2$ to produce that much methane.

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

$C O_{2 (g)} + 4 H_{2 (g)} \to C H_{4 (g)} + 2 H_{2} O_{(g)}$ $CO_(2(g)) + color(red)(4)H_(2(g)) -> CH_(4(g)) + 2H_2O_((g))$

Notice that 1 mole of carbon dioxide reacts with $4$ $color(red)(4)$ 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 $C O_{2}$ $CO_2$ and of $H_{2}$ $H_2$ reacted

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

and

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

Science

Math

Humanities

... and beyond

Question #bba7c

1 Answer

Related questions

Impact of this question