Answer:
[tex]\large \boxed{\text{8.00 mol}}[/tex]
Explanation:
We will need a balanced chemical equation with masses, moles, and molar masses.
1. Gather all the information in one place:
Mᵣ: 18.02
2Na + H₂O ⟶ 2NaOH + H₂
m/g: 72.0
2. Moles of H₂O
[tex]\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}[/tex]
3. Moles of Na
The molar ratio is 2 mol Na/1 mol H₂O.
[tex]\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}[/tex]
