Pure oxygen is prepared by thermal decomposition of KClO3 according to

1.	Pure oxygen is prepared by thermal decomposition of KClO3 according to the equation:KClO3(s) \(\overset{\Delta}{\rightarrow}\) KCl(s) + \(\frac{3}{2}\) O2(g)Calculate the volume of oxygen gas liberated at STP by heating 12.25 g KClO3(s).
Answer» 2KClO₃(s) \(\overset{\Delta}{\rightarrow}\) 2 KCl(s) + 3O₂(g) Molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5 \(\because\) 245 g KClO₃ produce = 22.4 × 3 L of oxygen \(\therefore\) 12.25 g KClO₃ will produce = \(\frac{22.4\times3}{24.5}\) x 12.25 = 3.36 L of oxygen.

Pure oxygen is prepared by thermal decomposition of KClO3 according to the equation:KClO3(s) \(\overset{\Delta}{\rightarrow}\) KCl(s) + \(\frac{3}{2}\) O2(g)Calculate the volume of oxygen gas liberated at STP by heating 12.25 g KClO3(s).

Answer»

2KClO₃(s) \(\overset{\Delta}{\rightarrow}\) 2 KCl(s) + 3O₂(g)

Molar mass of KClO₃ = 39 + 35.5 + 3(16) = 122.5

\(\because\) 245 g KClO₃ produce = 22.4 × 3 L of oxygen

\(\therefore\) 12.25 g KClO₃ will produce = \(\frac{22.4\times3}{24.5}\) x 12.25

= 3.36 L of oxygen.