# How many moles of sodium hydroxide will be produced from the complete reaction of 15.0 g sodium in the reaction 2Na + 2H_2O -> 2NaOH + H_2?

Apr 27, 2016

$\text{Moles of sodium hydroxide = Moles of sodium}$

#### Explanation:

So all we have to do is calculate the molar quantity of sodium, which is the same as the molar quantity of sodium hydroxide. How do I know this? Because you have kindly included the balanced equation, which shows a $1 : 1$ equivalence between the metal and its hydroxide salt, and a $2 : 1$ equivalence between the metal and the gas it has produced.

$\text{Moles of sodium}$ $=$ $\frac{15.0 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.652 \cdot m o l$.

Given the stoichiometry of the reaction, clearly $0.652 \cdot m o l$ sodium hydroxide will result.

How many grams of sodium hydroxide will this represent? What is the molar quantity of dihydrogen that will result, and what volume will it occupy under conditions of $1 \cdot a t m$ and $298 \cdot K$?