Question

calculate the mass of potassium chlorate required to librate 6.72 litere of oxygen at STP

Answer 1

6.72 liters of oxygen is:

$\frac {6.72L}{22.4 mol/L} = 0.3 mol$ , here we assume that O behaves as an ideal gas and we take 1 mole of gas as 22.4mol/L.

By the reaction below molar ratio O₂:KClO₃ is 3:2

$2KClO_3 = 2KCl + 3O_2$

So, by stoichiometry of the reaction we can find moles of potassium chlorate:

$0.3 moles \cdot \frac {2moles}{3moles}=0.2 moles$

mass of potassium chlorate is:

$0.2 moles \cdot 122.5 g/mol = 24.5 g$