Question

`10*g` of carbon disulfide is combusted with `15.5*L` of oxygen gas, which is the reagent in excess?

Answer 1

We have `"dioxygen gas"` in excess by approx. `"17 equiv"`.

Explanation:

We need (i) a stoichiometric equation........

`CS_2(l) + 3O_2(g) rarr CO_2(g) + 2SO_2(g)`

And (ii), equivalent quantities of `CS_2(l)`, and `"dioxygen"`.

`"Moles of"` `CS_2=(10.0*g)/(76.14*g*mol^-1)=0.0131*mol`.

Now (depending on your syllabus), `1*mol` of `"Ideal Gas"` occupies `22.7*L` at `"STP"`. If we (REASONABLY) assume ideality, then we have `(15.5*L)/(22.7*L*mol^-1)=0.683*mol*"dioxygen gas"`. And clearly, we have a stoichiometric EXCESS of dioxygen gas.

And thus `"EXCESS"` `O_2=(0.683-3xx0.0131)*mol`

`-=0.644*mol`.

Note that `CS_2` is (i) VERY FLAMMABLE and volatile; and (ii) PEN AND INKS very badly.