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

Jun 16, 2017

We have $\text{dioxygen gas}$ in excess by approx. $\text{17 equiv}$.

#### Explanation:

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

$C {S}_{2} \left(l\right) + 3 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 S {O}_{2} \left(g\right)$

And (ii), equivalent quantities of $C {S}_{2} \left(l\right)$, and $\text{dioxygen}$.

$\text{Moles of}$ $C {S}_{2} = \frac{10.0 \cdot g}{76.14 \cdot g \cdot m o {l}^{-} 1} = 0.0131 \cdot m o l$.

Now (depending on your syllabus), $1 \cdot m o l$ of $\text{Ideal Gas}$ occupies $22.7 \cdot L$ at $\text{STP}$. If we (REASONABLY) assume ideality, then we have $\frac{15.5 \cdot L}{22.7 \cdot L \cdot m o {l}^{-} 1} = 0.683 \cdot m o l \cdot \text{dioxygen gas}$. And clearly, we have a stoichiometric EXCESS of dioxygen gas.

And thus $\text{EXCESS}$ ${O}_{2} = \left(0.683 - 3 \times 0.0131\right) \cdot m o l$

$\equiv 0.644 \cdot m o l$.

Note that $C {S}_{2}$ is (i) VERY FLAMMABLE and volatile; and (ii) PEN AND INKS very badly.