# When 10*g dihydrogen gas, and 32*g dioxygen gas react, which is the limiting reagent. And what mass of water will result?

Jan 26, 2017

You need (i) a stoichiometrically balanced equation...........

#### Explanation:

${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow {H}_{2} O \left(g\right)$

And (ii) equivalent quantities of dioxygen and dihydrogen.

$\text{Moles of dihydrogen}$ $=$ $\frac{10 \cdot g}{2.02 \cdot g \cdot m o {l}^{-} 1} \cong 5 \cdot m o l$.

$\text{Moles of dioxygen}$ $=$ $\frac{32 \cdot g}{32.0 \cdot g \cdot m o {l}^{-} 1} \cong 1 \cdot m o l$

Now, clearly, ${O}_{2}$ is the limiting reagent, and there is a stoichiometric excess of ${H}_{2}$. At most $2 \cdot m o l$ of ${H}_{2}$ react quantitatively with the $1 \cdot m o l$ of ${O}_{2}$, to give $2 \cdot m o l$ of ${H}_{2} O$, which represents a mass of $36 \cdot g$.

At given temperature and pressure what volume of $\text{dioxygen}$ will be required for a given volume of $\text{dihydrogen}$?