Sodium and water react according to the following equation. if 51.5 g of sodium are added to access water, how many liters of hydrogen gas are formed the stp ? show all work. 2na+2h2o—> 2naoh+h2

Answer is: volume is 25.08 liters.

Balanced chemical reaction: 2Na + 2H₂O → 2NaOH + H₂(g).

m(Na) = 51.5 g; mass of sodium.

n(Na) = m(Na) ÷ M(Na).

n(Na) = 51.5 g ÷ 23 g/mol.

n(Na) = 2.24 mol; amount of sodium.

From chemical reaction: n(Na) : n(H₂) = 2 : 1.

n(H₂) = 2.24 mol ÷ 2.

m(H₂) = 1.12 mol; amount of hydrogen gas.

V(H₂) = n(H₂) · Vm.

V(H₂) = 1.12 mol · 22.4 L/mol.

V(H₂) = 25.08 L; volume of hydrogen gas.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

it is a solid

explanation: