# What mass of dihydrogen gas results from the oxidation of 20*g sodium metal by a mass of 10*g of water?

May 29, 2017

Well first we write a stoichiometric equation, and get a bit over $1 \cdot g$ of ${H}_{2} \left(g\right)$

#### Explanation:

We represent the reaction by the following stoichiometric equation:

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

You can double this if you like. I think the arithmetic is easier if you include the $\frac{1}{2}$ coefficient.......

$\text{Moles of sodium} = \frac{20 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.870 \cdot m o l$

$\text{Moles of water} = \frac{10 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 0.555 \cdot m o l$.

Now, here (UNUSUALLY) water is the reagent in DEFICIENCY. And thus we assume that only $0.555 \cdot m o l$ of water reacts, and that some of the sodium metal remains unreacted - an unusual scenario.

And thus we get $\frac{0.555 \cdot m o l}{2}$ dihydrogen gas evolved, i.e. $\frac{0.555 \cdot m o l}{2} \times 2.02 \cdot g \cdot m o {l}^{-} 1 = 1.11 \cdot g$