The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually

Question

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)+5O2(g)+5NO(g)?CO2(g)+H2O(g)+5NO2(g)+2OH(g)
Suppose that an atmospheric chemist combines 145mL of methane at STP, 895mL of oxygen at STP, and 59.5mL of NO at STP in a 1.8?Lflask. The reaction is allowed to stand for several weeks at 275 K.

Questions:

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? 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?

Explanation / Answer

CH4(g)+5O2(g)+5NO(g) -----> CO2(g)+H2O(g)+5NO2(g) +2OH(g)
at STP, all gases contain 22.4L/mole, therefore, we can determine the number of moles of each gas used.
0.145L / 22.4L/mole = 0.00647moles CH4
0.895L / 22.4L/mole = 0.03996moles O2
0.0595L / 22.4L/mole = 0.002656moles NO

the limiting reactant = NO since this can only produce 0.00052 moles of CO2
0.002656moles NO yields 0.0005312moles CO2, 0.0005312moles H2O, 0.002656moles NO2 and 0.0010624moles OH, this is 100% of the reaction that could occur.
if 92% of NO is consumed, only 0.002444moles of NO is used up leaving 0.0002125moles NO
theoretically, the total number of products = 0.00478moles
92% of this yield = actual yield = 0.004398moles products

we know that 0.002444moles of NO are used, therefore, the moles of each gas used:
0.002444moles NO(5O2/5NO) = 0.002444moles O2 used and 0.0004888moles CH4 used
what's left of each reactant? NO = 92% used leaving 0.0002125moles, 0.002444moles O2 are used leaving 0.0375moles O2, 0.00048moles of CH4 are used, leaving 0.005981moles CH4
so, the species of gas in the container = all reactants and products
there are 0.004398moles of product present and there are 0.04369moles of reactants left
a total number of moles of gas present in the container = 0.04808moles
to find pressure, plug this number into the ideal gas law pv=nrt
solve for p = nrt/v
p = 0.04808moles x 0.0821L-atm/mole-K x 275K / 1.8L
p = 0.603atm

this the total pressure in flask

the partial pressure of each gas by multiplying the total pressure by the mole fraction of the gas.

species

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