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)part a)suppose that an atmospheric chemist combines 150 ml of methane at stp, 885 ml of oxygen at stp, and 59.5 ml of no at stp in a 2.2-l flask. the reaction is allowed to stand for several weeks at 275k. part b)if the reaction reaches 88.0 % of completion (88.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275k? if the reaction reaches 88.0 % of completion (88.0% of the limiting reactant is consumed), what are the partial pressures of each of the products in the flask at 275k? part c)what is the total pressure in the flask?

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

(A)

First let's identify the limiting reactant, we're given the volume of each gas at STP, so we can use PV=nRT to calculate the moles of each gas:

CH₄⇒ n = 1atm*0.150L/(0.082atm·L·mol⁻¹·K⁻¹*273K) = 6.70x10⁻³ mol CH₄O₂⇒ n = 1atm*0.885L/(0.082atm·L·mol⁻¹·K⁻¹*273K) = 0.0395 mol O₂NO⇒ n = 1atm*0.0595L/(0.082atm·L·mol⁻¹·K⁻¹*273K) = 2.66x10⁻³ mol NO

With 6.70x10⁻³ mol CH₄ we would need (5*6.70x10⁻³) 0.0335 mol NO, there's not enough NO to fully react with CH₄ or O₂, thus NO is the limiting reactant.

(B)

The amount of NO moles that reacted is

2.66x10⁻³ * 0.88 = 2.341x10⁻³ mol NO reactedRemaining moles NO= 2.66x10⁻³ - 2.341x10⁻³ = 3.190x10⁻⁴ mol NO

Then we determine the moles of CH₄ and O₂ that reacted:

CH₄⇒  2.341x10⁻³ mol NO * (1molCH₄/5molNO) = 4.682x10⁻⁴ mol CH₄ reacted

Remaining moles CH₄ = 6.70x10⁻³ - 4.682x10⁻⁴ = 6.232x10⁻³ mol CH₄

O₂⇒  2.341x10⁻³ mol NO * (5molO₂/5molNO) = 2.341x10⁻³ mol O₂ reacted

Remaining moles O₂ = 0.0395 - 2.341x10⁻³ = 0.0371 mol O₂

Now we use PV=nRT to calculate the partial pressure of each reactant after the reaction, using V=2.2L and STP:

CH₄⇒ P = 6.232x10⁻³mol*0.082atm·L·mol⁻¹·K⁻¹*273.16K/2.2L=6.345x10⁻³ atm

O₂⇒ P = 0.378 atm

NO ⇒ P = 3.248x10⁻³ atm

(B) Part 2

We calculate the amount of each product formed and again use PV=nRT to calculate P:

CO₂ ⇒ 4.682x10⁻⁴ mol CH₄ *(1molCO₂/1molCH₄)=4.682x10⁻⁴ mol CO₂

P = 4.767x10⁻³ atm

H₂O ⇒ 4.682x10⁻⁴ mol CH₄ *(1molH₂O/1molCH₄)=4.682x10⁻⁴ mol H₂O

P = 4.767x10⁻³ atm

NO₂ ⇒ 4.682x10⁻⁴ mol CH₄ *(5molNO₂/1molCH₄)=2.341x10⁻³ mol NO₂

P = 0.0238 atm

OH ⇒ 4.682x10⁻⁴ mol CH₄ *(2molOH/1molCH₄) = 9.364x10⁻⁴ mol OH

P  = 9.533x10⁻³ atm

(C)

The total pressure is the sum of all the partial pressures

P₊=PCH₄ + PO₂ + PNO + PCO₂ + PH₂O + PNO₂ + POH

P₊=0.430 atm

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