Question

6. (25 pts) There's a giant 188 L balloon filled with 0.3333 atm of O2(g) and 0.6667 atm Fr() at room temperature (298 K). The balloon can not burst but it's volume can fluctuate. a. How many mols of each, O2 and F3, are present in the balloon? (Use R=0.08206) b. The F, reacts with the O, to produce OF2(g). Assuming the reaction happens so fast the balloon's volume remains constant, initially, what is the new total pressure inside the balloon? F:(8) + -_02(g) → __OF_(g) (UNBALANCED)

Answer 1

@ ideal gas equation (OVE NRT) Pon = 0.3333 atm P (F) = 0.667 atm. (02) to, = n RXT { T = 298k го у R 0.08206 Lath mut't-) 0.833 atm x 188 2 = no x 0.082068 298k Latm muitat r. ) 0.333 × 1880 0.08206 X 298 62.604 - / -2.56 mole of) no = (02) 24.45 oxy geg, na 188 r. ? 0.667 at in x 0.08206 atm matt A x 298 t = 125.396 n (F2) = 0.6078 189 2 u. us 0.08 206 X 298 in(f2) = 5.12 mole/ot flusne. (2 F2 + 0₂ (9) of 2 It the reaction happen so fast and volume is constant then temperatwe will increase and [ P & T 7 so pressure will also increase

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