A mixture of methane and helium gases is maintained in a 9.38 L flask at a pressure of 1.49 atm and a temperature of 86 °C. If the gas mixture contains 4.22 grams of methane, the number of grams of helium in the mixture is
First, we need to calculate the moles of methane present in the mixture using the ideal gas law:
PV = nRT
n = PV/RT
n = (1.49 atm)(9.38 L) / (0.0821 L*atm/mol*K * 359 K)
n = 0.46 moles
Next, we need to calculate the moles of helium present in the mixture. Since the total moles of gas in the mixture is 0.46 (from methane calculation), the moles of helium can be calculated by:
Total moles - Moles of methane = Moles of helium
0.46 moles - (4.22 g / 16.05 g/mol) = Moles of helium
0.46 - 0.263 = 0.197 moles
Finally, we can calculate the mass of helium in the mixture:
Mass = Moles x Molar mass
Mass = 0.197 moles x 4 g/mol = 0.788 grams
Therefore, the gas mixture contains 0.788 grams of helium.