# How to calculate the volume of hydrogen which generates from 2g of sodium?

Jan 31, 2016

We specify room temperature, atmospheric pressure, and predict a volume of approx $2$ ${\mathrm{dm}}^{3}$ ${H}_{2} \left(g\right)$.

We also ASSUME that the $2.0$ $g$ of sodium metal has been dropped into a bucket of water, and the gas evolved is collected by displacement of water in a separate graduated container.

#### Explanation:

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

Given this stoichiometry, for every mole of sodium, one half a mole of dihydrogen is evolved.

Moles of $N a$ $=$ $\frac{2.0 \cdot \cancel{g}}{22.99 \cdot \cancel{g} \cdot m o {l}^{-} 1}$ $=$ $0.0870$ $m o l$.

Given the stoichiometry, $0.0435$ $m o l$ dihydrogen gas are evolved.

Now it is a fact, that at room temperature ($298$ $K$) and $1$ $a t m$, 1 mole of ideal gas has a volume of $25.4$ ${\mathrm{dm}}^{3}$. (These molar volumes are often given in a examination!)

So given the conditions stated and assuming ideality, $25.4$ ${\mathrm{dm}}^{3} \cdot m o {l}^{-} 1$ $\times$ $0.0870 \cdot m o l$ $=$ $2.2$ ${\mathrm{dm}}^{3}$ $=$ $2.2$ $L$ volume of dihydrogen gas.