Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is: XeF6 undergoes complete hydrolysis to give P and other products. P then undergoes slow disproportionation in OH-/H2O to give products.
1
3
2
0
Solution:
XeF6+3H2O→XeO3+6HF ↓OH-/H2O(disproportionation) XeO3 + 2OH- → XeO42- + H2O XeO42- → XeO4(s)+ Xe(g) + O2(g) Therefore, two gases, Xe(g) and O2(g) are released in the final step.