CH_{4}(g) + 5 O_{2}(g) + 5 NO(g) → CO_{2}(g) + H_{2}O(g) + 5 NO_{2}(g) + 2 OH(g)

Calculate the initial moles of each reactant:

STP: P = 1 atm, T = 0°C (273.15 K)

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

${\mathbf{n}}_{{\mathbf{CH}}_{\mathbf{4}}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{}\overline{)\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{160}\mathbf{\hspace{0.17em}}\overline{)\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}\mathbf{)}}$

**n _{CH4} = 7.1382x10^{-3} mol**

${\mathbf{n}}_{{\mathbf{O}}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{}\overline{)\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{880}\mathbf{\hspace{0.17em}}\overline{)\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}\mathbf{)}}$

**n _{O2} = 0.03925 mol**

${\mathbf{n}}_{\mathbf{NO}}\mathbf{=}\frac{\mathbf{(}\mathbf{1}\mathbf{}\overline{)\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{0575}\mathbf{\hspace{0.17em}}\overline{)\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\overline{)L}\xb7\overline{)\mathrm{atm}}}{\mathrm{mol}\xb7\overline{)K}}}\mathbf{)}\mathbf{(}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}\mathbf{)}}$

**n _{NO} = 2.5653x10^{-3} mol**

From the balanced reaction 1 mol of CH_{4}(g) react with 5 moles of O_{2}(g) and 5 moles of NO(g)

Divide to moles of O_{2}(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 O _{2} = 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 CH_{4}(g) left = 7.1382x10^{-3} mol - (2.36x10^{-3} mol/5)

moles CH_{4}(g) left = 7.1382x10^{-3} mol - 4.72x10^{-4} mol**moles CH _{4}(g) left = 6.66x10^{-3} mol**

moles O_{2}(g) left = 0.03925 mol - 2.36x10^{-3} mol**moles O _{2}(g) left =**

moles NO(g) left = 2.5653x10^{-3} mol - 2.36x10^{-3} mol**moles NO(g) left = 2.05x10 ^{-3}**

Calculate pressure using the ideal gas equation:

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:

CH_{4}(g) + 5 O_{2}(g) + 5 NO(g) → CO_{2}(g) + H_{2}O(g) + 5 NO_{2}(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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