Answer:
n_{H_2O}=2.43molH_2O
Explanation:
Hello,
The carried out reaction is:
[tex]2Na+2H_2O-->2NaOH+H_2[/tex]
Thus, the moles of water that react with 56.0 g of sodium are computed as follows:
[tex]n_{H_2O}=56.0gNa*\frac{1molNa}{23gNa}*\frac{2molH_2O}{2molNa}\\n_{H_2O}=2.43molH_2O[/tex]
Best regards.