# Question #19875

May 18, 2017

3.74

#### Explanation:

$N {a}_{2} O \to 2 N a + O$ doesn't quite work because oxygen always exists as a diatomic molecule, so the equation would have to be $2 N {a}_{2} O \to 4 N a + {O}_{2}$.
In the question, there are 1.87 moles of Sodium Oxide so following our ratio there will be $2 \times 1.87 = 3.74$ moles