How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o

Question

2 years ago

10

f the work used to solve this problem.
2K + F2 yields 2KF

2 answers:

Sladkaya [172] · Answer 1 · 2022-04-26T06:47:28+03:00

Potassium 23.5g/39.0983g/mol = 0.601mol

The Ratio of reactants is 2 to 1 so (0.601mol)/2 = 0.3005mol

Therefore 0.3005mol of F2 is needed to find liters use

formula V = nRT/P (V)Volume = 22.41L

(T)Temperature = 273K or 0.0 Celsius

(P)Pressure = 1.0atm

<span>(R)value is always .08206 with atm n = 0.3005moles (273)(.08206)(0.3005)/1 = V V = 6.7319 Liters</span>

givi [52] · Answer 2 · 2022-04-26T08:11:08+03:00

givi [52]2 years ago

6 0

Hope it cleared your doubt