0.400 g of methane, 0.100 g of hydrogen gas, and 0.400 g of oxygen gas are put i
ID: 713868 • Letter: 0
Question
0.400 g of methane, 0.100 g of hydrogen gas, and 0.400 g of oxygen gas are put into a 10.0 L container at 25 °C. Calculate the partial pressure of each gas and the total pressure of the gas mixture. Calculate the total pressure if the methane reacted with the oxygen until one of them is exhausted (assume the water formed is a liquid and doesn't contribute to the total pressure.) Calculate the most probable speed of each type of molecule in the original mixture. What temperature would the oxygen need to be at in order to have the same Vmp as hydrogen at 25 °C? (10 pts)Explanation / Answer
moles of CH4 = 0.400g / 16g/mol = 0.025 mol
moles of H2 = 0.100g / 1.0079g/mol = 0.0992 mol
moles of O2 = 0.400g / 32g/mol = 0.0125 mol
Total moles, n = 0.025 + 0.0992 + 0.0125 = 0.1367 mol
Ptotal = nRT / V = (0.1367 mol * 0.0821 L atm mol-1K-1*298.15K) / 10L = 0.3346 atm
Now find mole fractions,
XCH4 = 0.025mol / 0.1367mol = 0.18288 ; XH2 = 0.0992mol /0.1367mol = 0.72567
XO2 = 0.0125mol / 0.1367mol = 0.09144
use Dalton's law to find partial pressures of gases
PCH4 = XCH4 * Ptotal = 0.18288*0.3346 = 0.06119 atm
PH2 = 0.72567*0.3346 = 0.24281 atm
PO2 = 0.09144*0.3346 = 0.03059 atm
CH4 + 2O2 ------> CO2 + 2H2O
Clearly CH4 is the limiting reactant in this reaction.
moles of CH4 = 0.025 mol
moles of O2 reacted = 2*0.025 = 0.05 mol
From the reaction stoichiometry it is clear that 1 mole of CH4 produces 1 mole of CO2 hence,
moles of CO2 formed = 0.025 mol
ntotal = 0.025+0.05+0.025 = 0.1 mol
Ptotal = (0.1mol*0.0821 L atm mol-1 K-1*298.15K) / 10L = 0.24478 atm
Total pressure when methane reacted with oxygen = 0.24478 atm.
Most probable speeds of each type of molecule in the original mixture,
vmp,CH4 = [2RT / m]1/2 = [2*8.314J/ mol K *298.15 K/ 0.400g]1/2= 111.3287 m/sec
vmp,H2 = [2*8.314*298.15 / 0.100]1/2 = 222.6575 m/sec
vmp,O2 = [2*8.314*298.15 / 0.400]1/2 = 111.3287 m/sec
[2*8.314*T / 0.400]1/2 = 222.6575 m/sec
(2*8.314*T) = (222.6575)2 * 0.400
T = 1192.6 K = 919.45oC
Hence, oxygen should be at 919.45oC in order to have the same vmp as hydrogen at 25oC
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