# What mass in grams of sodium hydroxide is produced if 20.0 g of sodium metal reacts with excess water according to the chemical equation 2Na(s) 2H_2O(l)->2NaOH(aq)+H_2(g)?

$N a \left(s\right) + {H}_{2} O \left(l\right) \rightarrow N a O H \left(a q\right) + \frac{1}{2} {H}_{2} \left(g\right)$
Moles of NaOH $=$ $\frac{20.0 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ ?? $m o l$
Clearly, there is a 1:1 equivalence between the moles of sodium, and moles of sodium hydroxide. One half an equiv of dihydrogen gas is also produced. We know that $1$ $m o l$ of gas occupies $24.5$ $L$ at $298$ $K$, and $1$ $a t m$. What volume of dihydrogen would be evolved under the above conditions?