What volume does 22.4 g of chlorine gas occupy at STP?

1 Answer
Jun 15, 2016

Answer:

#7.08 L#

Explanation:

Because we are at STP, we must use the ideal gas law equation
#PxxV = nxxRxxT#.

  • P represents pressure (must have units of atm)
  • V represents volume (must have units of liters)
  • n represents the number of moles
  • R is the proportionality constant (has units of #(Lxxatm)/ (molxxK)#)
  • T represents the temperature, which must be in Kelvins.

Next, list your known and unknown variables. Our only unknown is the volume of #Cl_2(g)#. Our known variables are P,n,R, and T.

At STP, the temperature is 273K and the pressure is 1 atm. The proportionality constant, R, is equal to 0.0821 #(Lxxatm)/ (molxxK)#
The only issue is the mass of #Cl_2(g)#, we need to convert it into moles of #Cl_2(g)# in order to use the formula.
#22.4cancel"g" xx (1molCl_2)/(70.90cancel"g")# = 0.316 mol #Cl_2#

Now all we have to do is rearrange the equation and solve for V like so:
#V = (nxxRxxT)/P#
#V = (0.316molxx0.0821(Lxxatm)/(molxxK)xx(273K))/(1atm)#
#V = 7.08 L#