# How would you determine the mass of hydrogen produced when 0.25 g of sodium reacts with water? Sodium hydroxide is the other product formed in the reaction.

May 18, 2017

Well, our priority is to write a stoichiometric equation..............

#### Explanation:

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

Given the stoichiometry, HALF AN EQUIV dihydrogen gas is evolved from ONE EQUIV natrium metal.

$\text{Moles of Na = } \frac{0.25 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1} = 0.0109 \cdot m o l$ WITH RESPECT TO the metal...........

Stoichiometry demands that $\frac{0.0109 \cdot m o l}{2}$ $\text{dihydrogen gas}$ is evolved, i.e. $5.44 \times {10}^{-} 3 \cdot m o l$, a mass of $11 \cdot m g$.