HardChemistryHigh School (11–12)The p-Block Elements › Groups 17 and 18: Halogens and Noble GasesNumericalConceptualNEETA-LevelIBAEEE (Amrita)AMUEEEAP EAPCET
A sample of 24.53 g of XeF6 (245.3 g/mol) is hydrolysed completely to XeO3 (179.3 g/mol) and…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
A sample of of XeF6 () is hydrolysed completely to XeO3 () and hydrogen fluoride. Find the mass of XeO3 and the moles of HF. Then find the mass of XeOF4 () if the same sample is hydrolysed only as far as . In the equation the xenon atoms that go to 0 gain electrons, and the oxygen of water that becomes O2 loses them. Find the electrons gained by the 4 xenon atoms that reach 0, and the electrons lost by the oxygen atoms that form 3 molecules of O2.
Need a reminder? The key idea
The worked solution
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Moles. of XeF6.
Step 2. Complete hydrolysis. gives of XeO3, which is , and of HF.
Step 3. Partial hydrolysis. One mole of XeOF4 for each mole of XeF6: .
Step 4. Electrons gained. Four xenon atoms go from to 0, so electrons are gained.
Step 5. Electrons lost. Three O2 molecules have 6 oxygen atoms, each going from to 0 and losing 2 electrons: . The other 4 electrons are lost by the 2 xenon atoms that go from to : . The totals match: .
Answer: of XeO3, of HF, of XeOF4, 16 electrons gained and 12 lost by oxygen.
Check. Electrons gained equal electrons lost, . The reaction is a disproportionation of xenon.
Discussion (0)
Log in to join the discussion.
Loading…