if 183.7g of KCLO3 is completly burnt catalytically, what volume of oxygen gas will be obtained at 39 degree and 1200 torr pressure

Question

its ainorganic chemistry numerical.

Answer 1

In the presence of a catalyst the balanced equation for combustion is

2 KClO₃(s) → 3 O₂(g) + 2 KCl(s)

moles KClO₃ used = 183.7 g x 1 mole/122.6 g = 1.498 moles

moles O₂ generated = 1.498 moles KClO₃ x 3 moles O₂/2 moles KClO₃ = 2.248 moles O₂ produced

Volume of O₂ produced equivalent to 2.248 moles:  Use PV = nRT

V = nRT/P = (2.248)(62.36)(312)/1200  (NOTE: use the correct value of R, and convert ºC to ºK)

V = 36.45 liters