Question #4e869

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Question #4e869

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Answer 1 · 2015-05-17T09:09:40

You'd need ${0.250 m}^{3}$ $"0.250 m"^3$ of carbon dioxide to react with that many cubic meters of hydrogen.

Start with the balanced chemical eqaution for your reaction, which is known as 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 you have a $1 : 4$ $1:color(red)(4)$ mole ratio between carbon dioxide and hydrogen, which means that, regardless of how many moles of the former react, you'll always use 4 times more moles of hydrogen gas.

Assuming both gases are under the same conditions for pressure and temperature, the mole ratio will become a volume ratio, i.e. regardless of haw many cubic meters of $C O_{2}$ $CO_2$ react, you'll always need 4 times more cubic meters of hydrogen for the reaction to take place.

Since you know how many cubic meters of hydrogen you have, you can use the volume ratio to determine how many cubic meters of $C O_{2}$ $CO_2$ reacted

$1 m^{3} H_{2} \cdot \frac{{1 m}^{3} C O_{2}}{4 m^{3} C O_{2}} = {0.250 m}^{3} C O_{2}$ $1cancel("m"^3H_2) * ("1 m"^3CO_2)/(color(red)(4)cancel("m"^3CO_2)) = color(green)("0.250 m"^3CO_2)$

To determine how many cubic meters of methane you'd produce, use the fact that 1 mole, or 1 cubic meter, of $C O_{2}$ $CO_2$ produces exactly 1 cubic meter of $C H_{4}$ $CH_4$ ( $1 : 1$ $1:1$ mole ratio that exists between the two compounds)

This means that you'd get

$0.250 m^{3} C O_{2} \cdot \frac{{1 m}^{3} C H_{4}}{1 m^{3} C O_{2}} = {0.250 m}^{3} C H_{4}$ $0.250cancel("m"^3 CO_2) * ("1 m"^3CH_4)/(1cancel("m"^3CO_2)) = color(green)("0.250 m"^3CH_4)$

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Question #4e869

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