Question

If you start with 56.0 grams of sodium, how many moles of water will react with it?

Answer 1

Answer is: 2.43 mol of water.
Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂.
m(Na) = 56 g.
n(Na) = m(Na) ÷ M(Na).
n(Na) = 56 g ÷ 23 g/mol.
n(Na) = 2.43 mol.
From chemical reaction: n(Na) : n(H₂O) = 2 : 2 (1 : 1).
n(H₂O) = 2.43 mol.
m(H₂O) = 2.43 mol · 18 g/mol.
m(H₂O) = 43.82 g.

Answer 2

n_{H_2O}=2.43molH_2O

Step-by-step explanation:

Hello,

The carried out reaction is:

`2Na+2H_2O-->2NaOH+H_2`

Thus, the moles of water that react with 56.0 g of sodium are computed as follows:

`n_(H_2O)=56.0gNa*(1molNa)/(23gNa)*(2molH_2O)/(2molNa)\\n_(H_2O)=2.43molH_2O`

Best regards.