# Question #89b68

Jan 31, 2018

$4 N a + {O}_{2} \to 2 N {a}_{2} O$

Mass of 1 mole of sodium is 22.99 g. Therefore, 9.5 g is equivalent to $\frac{9.5}{22.99}$ moles., that is, 0.413 mole of sodium.

4 moles of Na require 1 mole of ${O}_{2}$.
0.413 mole of Na would therefore require $\left(\frac{1}{4}\right) \cdot 0.413$ moles of ${O}_{2}$.

Solving this, we get,

0.1033 moles of oxygen is needed to completely react with 9.5 g of Na.