Question

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You have 2.2 mol Xe and 1.8 mol F2, but when you carry out the reaction you end up with only 0.25 mol XeFa. What is the percent yield of this experiment? Xe(g) + 2 F2 (g) + XeF4 (g) Attempts remaining: 5 If 10.0 moles of O2 are reacted with excess NO in the reaction below, and only 7.8 mol of NO2 were collected, then what is the percent yield for the reaction? 2 NO (g) + O2 (g) → 2 NO2 (g) Attempts remaining: 5 How many moles of NO2 would be produced from 7.8 mol of O2 in the reaction below assuming the reaction has a 57.0% yield? 2 NO (g) + O2 (g) → 2 NO2 (g) Attempts remaining: 5

Answer 1

Balamced chomi cal eauation Xe) Fag) eF ) moles Fiu cue mole XeFos018 mol Fgives limitivg eaaut. e to0 O:9 mol aud ateual gield i'3025 ACtaad yiekd Theonen'cal yield .9 md Ke Fo , teure F ical yield af X iR mol 1007 Pers cent yidld POYCEN+ yield Ballanced clumial eguah'ou NOz) aNO@ Larm Fsom tae balaured eguakou, Em oue mole Oz R molos No2. teUce 10 moleg 0g dosm ao molee Noz . Se Hha tuooreti cal yield i ao moke cud the actual yilld i78mol Pencent yilld = ao Pentent Yild =39 %. Theoneti cal illd NOfoom mo 2 i -8X2 5.6 moles No fomed calculated as PETcent yidld=5x- Pentent yield x Teorèti cal yids Atual yield 57.0 xIS-6 dual gield farmed 8.9md NOR