18gr of CaH2 reacted with a small amount of water. After the reaction, the mass of the solid is 30g.

18gr of CaH2 reacted with a small amount of water. After the reaction, the mass of the solid is 30g. Determine the mass fraction of oxygen in the solid.

Given:
m (CaH2) = 18 g
m (tv.in-va) = 30 g

To find:
ω (O in tv. in-ve) -?

Decision:
1) CaH2 + 2H2O => Ca (OH) 2 + 2H2 ↑;
2) Let m react. (CaH2) = (x) g, m rest. (CaH2) = (18 – x) d;
3) n react. (CaH2) = m reag. (CaH2) / M (CaH2) = (x / 42) mol;
4) n (Ca (OH) 2) = n reag. (CaH2) = (x / 42) mol;
5) m (Ca (OH) 2) = n (Ca (OH) 2) * M (Ca (OH) 2) = ((x / 42) * 74) mol;
6) m (solid) = m rest. (CaH2) + m (Ca (OH) 2);
30 = (18 – x) + ((x / 42) * 74);
x = 15.75;
7) n (Ca (OH) 2) = x / 42 = 15.75 / 42 = 0.375 mol;
8) n (O) = n (Ca (OH) 2) * N (O) = 0.375 * 2 = 0.75 mol;
9) m (O) = n (O) * M (O) = 0.75 * 16 = 12 g;
10) ω (O) = m (O) * 100% / m (tv.in-va) = 12 * 100% / 30 = 40%.

Answer: Mass fraction of O is 40%.