Answer:
258.71 dm³
Step-by-step explanation:
We'll begin by calculating the number of mole of O₂ produced by the decomposition of 12 moles of KClO₃. This can be obtained as follow:
The balanced equation for the reaction is given below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 12 moles of KClO₃ will decompose to produce = (12 × 3)/2 = 18 moles of O₂.
Finally, we shall determine the volume of the O₂. This can be obtained as follow:
Temperature (T) = 325 K
Pressure (P) = 188 KPa
Number of mole (n) = 18 moles
Gas constant (R) = 8.314 KPa.dm³/Kmol
Volume (V) =?
PV = nRT
188 × V = 18 × 8.314 × 325
188 × V = 48636.9
Divide both side by 188
V = 48636.9 / 188
V = 258.71 dm³
Thus, 258.71 dm³ of oxygen were obtained from the reaction.