A30.0 mole% co and 70.0 mole% n2 gas mixture is stored in a cylinder with a volume of 3.5 ft3 at a temperature of 85°f. the reading on a bourdon gauge attached to the cylinder is 500 psi. calculate the total amount of gas (lb-mole) and the mass of co (lbm) in the tank.

Total gas = 0.308 lb-moles
 Mass CO = 2.59 lbm
   The ideal gas law is expressed by:
 PV = nRT
 where
 P = Pressure
 V = Volume
 n = number of moles of gas particles
 R = Ideal gas constant
 T = Absolute temperature
   Since finding the Ideal gas constant is far easier using the metric system, let's first convert all the input units to their appropriate metric units and then we'll convert the output to the imperial units desired.
   3.5 cubic feet is 3.5 ft^3 * 0.0283168 m^3/ft^3 = 0.0991088 m^3
 85°F = (5/9)(85-32) + 273.15 = 302.5944444 K
 500 psi = 500 * 6894.76 = 3447380 Pascals
 But a Bourdon gauge does not measure absolute pressure, it measures pressure relative to the ambient pressure. Since none was specified, I'll assume an ambient pressure of 1 atmosphere. So we need to adjust the pressure by that amount:
 3447380 Pa + 101325 Pa = 3548705 Pa
   For the units used, the ideal gas constant is:
  8.3144598 m^3*Pa/(K*mol)
   So let's solve the formula for n, substitute the known values, and calculate: PV = nRT PV/(RT) = n 3548705 Pa * 0.0991088 m^3/(8.3144598 m^3*Pa/(K*mol) * 302.5944444 K) = n 351707.8941 Pa*m^3/(2515.909344 Pa*m^3/mol) = n
 139.7935482 mol = n
   So we have 139.7935482 moles of gas particles in the cylinder. Now to convert to the imperial units.
   A lb-mol is defined as the number of atoms in 12 pounds of carbon-12, just like a mole is defined as the number of atoms in 12 grams of carbon-12. So 1 lb-mol is 453.59237 moles. So 139.7935482 mol / 453.59237 mol/lb-mol = 0.308192019 lb-mole
   Since we have 30% of that being CO, the amount of CO is
 0.308192019 lb-mole * 0.30 = 0.092457606 lb-mol
   Mass of CO
  (12 + 16) * 0.092457606 = 2.588812957 lbm
   Rounding to 3 significant figures gives:
 Total gas = 0.308 lb-moles Mass CO = 2.59 lbm

Elements would be the right answer