30 moles of oxygen were admitted to the ozonizer and as a result, 25 moles of oxygen and ozone came out together.

30 moles of oxygen were admitted to the ozonizer and as a result, 25 moles of oxygen and ozone came out together. Determine the mass fraction of ozone in the gas mixture.

Given:
n total (O2) = 30 mol
n mixture (O2, O3) = 25 mol

Find:
ω (O3) -?

Solution:
1) 3O2 => 2O3;
2) Let n react. (O2) = (x) mol, then n sample. (O3) = (x * 2/3) mol;
3) 25 = (30 – x) + (2x / 3);
75 = 90 – 3x + 2x;
x = 15;
n react. (O2) = x = 15 mol;
4) n rest. (O2) = n total (O2) – n react. (O2) = 30-15 = 15 mol;
5) m rest. (O2) = n rest. (O2) * M (O2) = 15 * 32 = 480 g;
6) n (O3) = 2х / 3 = 2 * 15/3 = 10 mol;
7) m (O3) = n (O3) * M (O3) = 10 * 48 = 480 g;
8) m mixture (O2, O3) = m rest. (O2) + m (O3) = 480 + 480 = 960 g;
9) ω (O3) = m (O3) * 100% / m mixture (O2, O3) = 480 * 100% / 960 = 50%.

Answer: Mass fraction of O3 is 50%.