Na finally exists in the solution in the form of NaOH, which is neutral after reacting with acid.

That is, NNA+= NOH-= NH+= 0.08 * 0.5 * 2 = 0.08 mole.

mNa=nNa*M=0.08*23= 1.84g

2

This is to test the familiarity of the reaction. If you are not familiar with it, you can also calculate that it contains not only Na and Na2O, but also Na2O.

The generated gas is 0.336L, which can be completely electrically burned without gas residue.

2H2 - O2

2 1

Here VH2:VO2=2: 1=nH2:nO2.

NH2 = 0.336/3 * 2/22.4 = 0.0 1 mol.

nO2 = 0.336/3 * 1/22.4 = 0.005mol

2Na+2H2O=2NaOH+H2↑

2 1

X 0.0 1 mol

X = 0.02mol mol

mNa=0.02*23=0.46g

2Na2O2+2H2O=4NaOH+O2↑

2 1

Y 0.005mol mol

Y = 0.005mol mol

MNa2O2 = 0.005 * 78 = 0.39g

nNa = 0.02+0.005 * 2 = 0.03mol & lt； 0.08mol, so the reactant has other substances.

(Check equations and inferences here. You can write directly if you are familiar with it. If you are not familiar with it, you can imagine that apart from Na and Na2O, only compounds of Na and O exist, or you can write directly.

4Na+O2=2Na2O

2Na2O+O2=2Na2O2)

Namely Na2O

According to the conservation law of sodium,

Na2O contains nNa=0.08-0.03=0.05mol.

That is, nna2o = 0.05/2 = 0.025mol.

mNa2O=0.025*62= 1.55g

So the remaining solid components and mass are respectively

Sodium: 0.46 g.

Sodium oxide: 0.39g.

Na2O:1.55g.