anonymous
  • anonymous
How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all of the work used to solve this problem. 2 K + F2 yields 2 KF
Chemistry
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Potassium (K) = 39.098 g/mol 23.5g / 39.098 g/mol = 0.601 mol ratio: 2:1 0.601 mol / 2 = 0.3005 mol V = nRT/P = 22.4 L - one mole of any gas occupies a volume of 22.4 L n = 0.3005 mol R = 0.0821 T = 273.15 K P = 1 atm (0.3005 * 0.0821 * 273.15) / 1 = 6.73 V = 6.73 L
anonymous
  • anonymous
Can someone tell me if my answer is correct, please?
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  • anonymous

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anonymous
  • anonymous
Cuanchi
  • Cuanchi
yes its correct!
anonymous
  • anonymous
Hey, I'm back, I was away from my computer for a while. Thanks! :)

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