In methane combustion, if $2.5 \cdot m o l$ $2.5*mol$ $O_{2}$ $O_2$ were present, what molar quantity of methane would react?

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In methane combustion, if $2.5 \cdot m o l$ $2.5*mol$ $O_{2}$ $O_2$ were present, what molar quantity of methane would react?

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Answer 1 · 2016-03-30T07:07:38

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ $CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)$ . So $1.25$ $1.25$ $m o l$ $mol$ methane are necessary for the given quantity of dioxygen.

Explanation:

$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ $CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)$ .

This equation clearly tells us that each mole of methane requires 2 mole of dioxygen for complete combustion. There are $2.5$ $2.5$ $m o l$ $mol$ $O_{2}$ $O_2$ ; if complete combustion occurred, then $1.25$ $1.25$ $m o l$ $mol$ methane gas were combusted.

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In methane combustion, if $2.5 \cdot m o l$ $2.5*mol$ $O_{2}$ $O_2$ were present, what molar quantity of methane would react?

1 Answer

Explanation:

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In methane combustion, if $2.5 \cdot m o l$ $2.5*mol$ $O_{2}$ $O_2$ were present, what molar quantity of methane would react?