Question

Pt arderea a 2 moli de hidrocarbura ce contine 85,714% C se consuma 268,8 L de o2. Care este formula moleculara a hidrocarburii?

Answer 1

%H = 100 - 85.714 = 14.286%
85.714/12 = 7.14333
14.286/1 = 14.826
Imparti la 7.1433 => FB: CH2
(CH2)n + 3n/2 O2 => n CO2 + n H2O
1 mol hidrocarbura........3n/2 * 22.4 l oxigen
2 moli.............................268.8
3n*22,4 = 268.8 => 3n = 12 => n = 4
FM: C4H8