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?

Calculate the initial moles of each reactant:

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

$$\begin{gathered} \boxed{\mathbf{PV}\mathbf{=}\mathbf{nRT}} \\ \mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}} \end{gathered}$$

${\mathbf{n}}_{{\mathbf{CH}}_{\mathbf{4}}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{160}\mathbf{\hspace{0.17em}}\cancel{\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\cancel{L}·\cancel{\mathrm{atm}}}{\mathrm{mol}·\cancel{K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{K}\mathbf{)}}$

n_CH4 = 7.1382x10^-3 mol

${\mathbf{n}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{880}\mathbf{\hspace{0.17em}}\cancel{\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\cancel{L}·\cancel{\mathrm{atm}}}{\mathrm{mol}·\cancel{K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{K}\mathbf{)}}$

n_O2 = 0.03925 mol

${\mathbf{n}}_{\mathbf{NO}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{ }\cancel{\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0575}\mathbf{\hspace{0.17em}}\cancel{\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\cancel{L}·\cancel{\mathrm{atm}}}{\mathrm{mol}·\cancel{K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{K}\mathbf{)}}$

n_NO = 2.5653x10^-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:

$\mathbf{moles}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{0}\mathbf{.}\mathbf{03925}\mathbf{ }\mathbf{mol}}{\mathbf{5}}$

moles O₂ = 7.852x10^-3 mol

$\mathbf{moles}\mathbf{ }\mathbf{NO}\mathbf{=}\frac{\mathbf{2}\mathbf{.}\mathbf{5653}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{mol}}{\mathbf{5}}$

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: