A sample of 24.53 g of XeF6 (245.3 g/mol) is hydrolysed completely to XeO3 (179.3 g/mol) and…
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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 ideaShow
Xenon has 8 outer electrons, and each fluorine atom uses one of them. A double bond to oxygen counts as one region and uses 2 electrons. XeF2, XeF4 and XeF6 are made from Xe and F2 in the ratios 1 : 1, 1 : 2 and 1 : 3. The hydrolysis of XeF6 gives XeO3 and HF. In a disproportionation the same element is oxidised and reduced.
The worked solution
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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.
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