How many grams of gas are present in a sample that has a molar mass of 70.0 g/mol and occupies a 2.00 l container at 117 kpa and 35.1 degrees celsius?

The ideal gas law of gases states that the volume occupied by certain moles  of a gas has a pressure at a temperature in kelvin. The law combines both boyle's law and the charle's law. Therefore, PV=nRT where P is the pressure, V is the volume, n is the number of moles, T is the temperature and R is the gas constant. 1 atm = 101.325 kPa, therefore; 117kPa = 1.155 atm , 35.1 °c = 308.25 Kelvin and R = 0.0821
PV=nRT
n = PV/RT
   = (1.155 × 2)/ (0.0821 × 308.25)
  =  0.0913 moles
But, 1 mole = 70 g
Therefore, 0.0913 moles × 70 = 6.39 g

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