Question

For many years, it was thought that no reaction could produce sodium perbromate, but the dis- covery of a reaction producing the equally elu- sive xenon difluoride, XeF2, led to the discovery of the following reaction that yields sodium perbromate. (OBJECTIVES 2 & 3) NaBroz + XeF2 + H20 + NaBrO4 + 2HF + Xe a. Write a conversion factor that could be used to convert between moles of xenon difluoride, XeF2, and moles of hydrogen fluoride, HE b. How many moles of XeF, are necessary to form 16 moles of hydrogen fluoride? c. What is the maximum number of moles of NaBrO, that could form in the combina- tion of 2 moles of NaBrO3 and 3 moles of XeF? d. What is the maximum number of moles of NaBrO4 that could form in the combina- tion of 2 moles of NaBrO2 and 3 million moles of XeFZ? e. Write a conversion factor that could be used to convert between moles of sodium perbromate, NaBrO4, and moles of hydro- gen fluoride, HF.

Answer 1

+ + NbS10a Imola x Imol. 1,0 Imol. . » Nh SAO + GHE + Xe. >ê) Imol. anol. Imol. @ Gruversion facte to Convert mole of Xef to HF moder Moles of HF & mol. xſmoke of XF) I mol. Foogy ewди Ф, I moll moles of Xe fa fresked - (moles of HF) = (16) moles +8 melep ҒАси ееји (В), , Cruciting Treactant is NaBros, here Imole of NaBroz zivel mole of NaBro4 = Imol. ä mole- of NaBn03 gives moles of Nabilly - tla) mol - a molt Saule as ©. Fious con modes of KARTA = HF I'mol. (moles of Mariyogi). → NABROG amol. I @ @ Farul evvu @, . Moles of HE frecued = a (mokes of Nar04) = 2(5.8aa) mol 11.644 mol