If Billy Bob bought you an “I Luv U” balloon holding 6.75 L of Helium, how many moles would Billy Bob have

Question

If Billy Bob bought you an “I Luv U” balloon holding 6.75 L of Helium, how many moles would Billy Bob have bought you

Answer 1

need more info

Answer 2

Assume STP (20C, 101.325 kPa) and use ideal gas law; i.e. pV=nRT; therefore

n=pV/RT

You are assuming the pressure and temperature (remember you need to state your assumptions in the answer, I used the 20C STP as it seems more places have an average temp of 20C compared with 0C), R is constant, substitute in the volume of 6.75.

You must remember to make sure that the units you use for temperature and pressure match the units of the gas constant (gas constant value depends on the units).  I generally use the 8.314 and in this case you need temperature in Kelvin (STP=293.15K) pressure in kPA (101.325), and volume in litres.  

Therefore:

n = pV/RT

   = (101.325kPa).(6.75L) / (8.314L.kPa.K^-1.mol^-1).(293.15K)

  = 0.28mol (if my arthimatic is correct-I'd double check it though)

Answer 3

We can only do this problem if we assume that the balloon is at STP. That is the helium inside the ballon has a pressure of 1 atmosphere and the temperature is 0°C.

Under these conditions, 1mol He will occupy a volume of 22.4litre. You have 6.75 litres, therefore you have:

6.74/22.4 = 0.300 mol

Answer 4

that would involve what volume a mole of Helium fills under standard temperature and pressure and what 6.75L corresponds to in moles.

Answer 5

6,75 L of He ------ x mol He

22,4 L          ------- 1 mol He

--------------------------------------...

x= 6,75L x 1 mol He / 22, 4 L

x= 0,301 mol He

1 mol --------- 6,023x10^23 molecules (atoms, ions, any)

0, 301--------- x

_______________________

x= 6,023x10^23  x 0,301 / 1

X= 1,81x10^23 molecules