Determine the mass fraction of the substance in the solution obtained after the interaction of 0.115 g

Determine the mass fraction of the substance in the solution obtained after the interaction of 0.115 g of sodium in 200 ml of a 10% solution of sodium cellulose with a density of 1.11 g ml.

Given:
m (Na) = 0.115 g
V (NaOH) = 200 ml
W (NaOH) = 10%
p = 1.11 g / ml
To find:
W (substances) =?
Solution:
1) Find the mass of the solution:
m (solution) = V x p = 220 ml x 1.11 g / ml = 222 g.
2) W = (m substance: m solution) x 100%
m (substance) = (222 g x 10%): 100% = 22.2 g.
3) We compose the reaction equation.
2Na + 2H2O = 2NaOH + H2
M (Na) = 23 x 2 = 46 g / mol
M (NaOH) = (23 + 16 + 1) x 2 = 80 g / mol
M (H2) = 1 x 2 = 2 g / mol
4) Let NaOH be x g, then: 0.115 g: 46 g / mol = x: 80 g = 0.2 g.
Let H2 be yr, then 0.115 g: 46 g / mol = x: 2 r = 0.005 g.
5) Find the mass of the resulting solution.
m solution = (222 + 0.115) – 0.005 = 222.11 g.
6) Find the mass of the resulting substance.
m substance = 22.2 + 0.2 = 22.4 g.
7) W = (22.4: 222.11) x 100% = 10.09%
Answer: W (substances) = 10.09%.