Question

A `30*L` volume of dioxygen gas at `473*K`, and a pressure of `2*atm`, will completely combust WHAT mass of `"ethane gas"`?

Answer 1

We need to work out (i) moles of oxygen gas, and (ii) equivalent moles of ethane.

Explanation:

`n=(PV)/(RT)` `=` `((2*atm)(30.0*L))/((0.0821*L*atm*K^(-1)*mol^-1)(473*K))`

`=` `1.54` `mol` `O_2(g)`.

Now, combustion reaction of ethane can be given as:
`C_2H_6(g) + 7/2O_2(g) rarr 2CO_2(g) + 3H_2O(g)`

So, this `1.54` `mol` quantity represents 7/2 equiv of ethane by the stoichiometry of the rxn. Thus moles of ethane = `1.54*molxx2/7` `=` `0.440` `mol` ethane.

So (finally!), mass of ethane `=` `0.440*molxx30.07*g*mol^(-1)` `=` `??g`. Approx. 12 g?