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Problem: The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reactions of a number of such steps for methane gas is as follows:CH4(g) + 5 O2(g) + 5 NO(g) → CO2(g) + H2O(g) + 5 NO2(g) + 2 OH(g)Suppose that an atmospheric chemist combines 160 mL of methane at STP, 880 mL of oxygen at STP, and 57.5 mL of NO at STP in a 2.2-L flask. The reaction is allowed to stand for several weeks at 275 K. If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275 K?

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FREE Expert Solution
CH4(g) + 5 O2(g) + 5 NO(g) → CO2(g) + H2O(g) + 5 NO2(g) + 2 OH(g)


Calculate the initial moles of each reactant:

STP:   P = 1 atm,   T = 0°C (273.15 K)

PV=nRTn=PVRT

nCH4=(1 atm)(0.160L)(0.08206L·atmmol·K)(273.15 K)

nCH4 = 7.1382x10-3 mol


nO2=(1 atm)(0.880L)(0.08206L·atmmol·K)(273.15 K)

nO2 = 0.03925 mol


nNO=(1 atm)(0.0575L)(0.08206L·atmmol·K)(273.15 K)

nNO2.5653x10-3 mol


From the balanced reaction 1 mol of CH4(g) react with 5 moles of O2(g) and 5 moles of NO(g)

Divide to moles of O2(g) and moles of NO(g) for stoichiometry:

moles O2=0.03925 mol5

moles O2 = 7.852x10-3 mol

moles NO=2.5653×10-3 mol5

moles NO = 5.1305x10-4 mol


NO(g) → limiting reactant → will be consumed totally, but only 92.0% reacted

moles NO(g) reacted = (92.0%) (2.5653x10-3 mol)
moles NO(g) reacted = 2.36x10-3 mol


moles CH4(g) left = 7.1382x10-3 mol - (2.36x10-3 mol/5)
moles CH4(g) left = 7.1382x10-3 mol - 4.72x10-4 mol
moles CH4(g) left = 6.66x10-3 mol

moles O2(g) left = 0.03925 mol - 2.36x10-3 mol
moles O2(g) left = 0.0369 mol

moles NO(g) left = 2.5653x10-3 mol - 2.36x10-3 mol
moles NO(g) left = 2.05x10-3 mol

Calculate pressure using the ideal gas equation:

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Problem Details
The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reactions of a number of such steps for methane gas is as follows:

CH4(g) + 5 O2(g) + 5 NO(g) → CO2(g) + H2O(g) + 5 NO2(g) + 2 OH(g)

Suppose that an atmospheric chemist combines 160 mL of methane at STP, 880 mL of oxygen at STP, and 57.5 mL of NO at STP in a 2.2-L flask. The reaction is allowed to stand for several weeks at 275 K. If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275 K?

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What scientific concept do you need to know in order to solve this problem?

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Based on our data, we think this problem is relevant for Professor Delgado's class at FIU.