# 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 products in the flask at 275 K?

###### 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)

$\overline{)\mathbf{PV}\mathbf{=}\mathbf{nRT}}\phantom{\rule{0ex}{0ex}}\mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}}$

nCH4 = 7.1382x10-3 mol

nO2 = 0.03925 mol

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 = 7.852x10-3 mol

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

From the balanced reaction 1 mole of CH4(g) produces 1 mole CO2(g), 1 mole H2O(g), 5 moles NO2(g), and 2 moles OH(g)

Calculate moles of CH4 reacted:

moles CH4(g) reacted = 4.72x10-4 mol

moles CO2(g) = 1 (4.72x10-4 mol)
moles CO2(g) = 4.72x10-4 mol

moles H2O(g) = 1 (4.72x10-4 mol)
moles H2O(g) = 4.72x10-4 mol

moles NO2(g) = 5 (4.72x10-4 mol)
moles NO2(g) = 2.36x10-3 mol

moles OH(g) = 2 (4.72x10-4 mol)
moles OH(g) = 9.44x10-4 mol

Calculate pressure using the ideal gas equation:

95% (390 ratings) ###### 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 products in the flask at 275 K?

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