Question

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)`?

Answer 1

Approx. thirty grams.

Explanation:

`"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`