Question

Question #fd6e0

Answer 1

Approx. `30*mg` `O_2` gas are required.

Explanation:

We need a stoichiometrically balanced equation:

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

Is this balanced? Why?

The equation tells us unequivocally that 1 mole of methane gas requires 2 moles of dioxygen gas for complete combustion.

`"Moles of methane"` `=` `(7.90xx10^-3*g)/(16.04*g*mol^-1)` `=4.93xx10^-4*mol`.

And thus 2 equiv of dioxygen gas are required (I use the term `"dioxygen"` to avoid confusion with regard to the number of oxygen atoms I need) per equiv of methane gas.

`"Moles of dioxygen"` `=` `4.93xx10^-4*cancel(mol)xx2xx32.00*g*cancel(mol^-1)=??g`