If 5.563 grams of $CH_4$ reacted with 25.00 grams of oxygen, how much $CO_2$ , in grams, can be produced? $CH_4(g) + 2O2(g) -> CO_2(g) + 2H_2O(g)$?

Md Ariful Islam

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Md Ariful Islam

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$"Moles of methane"$ $=$ $(5.563*g)/(16.043*g*mol^-1)$ $=$ $0.3468*mol$.

$"Moles of dioxygen"$ $=$ $(25.00*g)/(31.998*g*mol^-1)$ $=$ $0.7813*mol$.

Clearly there is sufficient dioxygen to react with methane as per the stoichiometric equation:

$CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)$

which requires 2 equiv dioxygen per equiv of methane. So we get 1 equiv of carbon dioxide, which has a mass of $2xx0.3468*molxx44.010*g*mol^-1$ $=$ $??g$

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