# How much dihydrogen would evolve if sodium metal were dropped into water?

Jul 18, 2017

You need to specify a mass of sodium metal.......

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

#### Explanation:

If $1 \cdot g$ of sodium metal were dropped into a beaker of water, along with spitz and sparx we would get......

$\frac{1}{2} \times \frac{1 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} \times 2.02 \cdot g \cdot m o {l}^{-} 1 = 0.044 \cdot g$ $\text{dihydrogen gas}$.

Of course, if we specify the temperature, and assume $1 \cdot a t m$ pressure, we could calculate the volume of dihydrogen gas produced.