The emission of NO_2 by ofssil fuel combustion can be prevented by injecting gas

Question

The emission of NO_2 by ofssil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea reduces NO (which oxidizes in air to ofrm NO_2) according to the ofllowing reaction: 2CO(NH_2)_2(g) + 4NO(g) + O_2(g) right arrow 4N_2(g) + 2CO_2(g) + 4H_2 O (g) Suppose that the exhaust stream of an automobile has a flow rate of 2.48 L/s at 662 K and contains a partial pressure of NO of 10.7 torr. What total mass of urea is necessary to react completely with the NO ofrmed during 7.5 hours of driving? Express your answer using two significant figures. Air conditioners not only cool air, but dry it as well. Suppose that a room in a home measure 7.0m times 10.0m times 2.3m. If the outdoor temperature is 30 degree C and the vapor pressure of water in the air is 80% of the vapor pressure of water at this temperature. what mass of water must be removed from the air each time the volume of air in the room is cycled through the air conditioner? The vapor pressure ofr water at 30 degree C is 31.8 torr. Express your answer using two significant figures.

Explanation / Answer

Solution :-

item 11 part A )

Vapor pressure of NO = 10.7 torr

Flow rate = 2.48 L /s

Time = 7.5 hour

T= 662 K

Lets calculate the total volume given out in 7.5 hour

(2.48 L /s) * (3600 s * 7.5 hr / 1 hr) = 66960 L

Now lets calculate the moles of the NO using the ideal gas law equation

PV= nRT

PV/RT = n

(10.7 torr *1 atm / 760 torr )* 66960 L / 0.08206 L atm per mol K * 662 K = n

17.35 mol = n

So the moles of NO given out are 17.35 mol

Now lets convert moles to mass of NO

Mass = moles * molar mass

          = 17.35 mol * 30.007 g per mol

         = 521 g NO

So the mass of NO produced = 521 g

Part 2)

Room dimensions are 7.0 m * 10.0 m*2.3 m

So the volume of room = 7.0 m * 10.0 m * 2.3 m = 161 m3

Lets convert it to liter

161 m3 * 1000 L / 1 m3 = 1.61*10^5 L

T = 30 C +273 = 303 K

Vapore pressure of H2O = 31.8 torr

Vapor pressure of water in air = 80 %

So lets calculate the pressure of water in the air

(31.8 torr * 1 atm / 760 torr)*(80 % /100%) = 0.033474 atm

Now lets calculate the moles of water

PV= nRT

PV/RT = n

0.033474 atm * 1.61*10^5 L / 0.08206 L atm per mol K * 303 K = n

219 mol = n

Now lets convert moles to mass of water

Mass= moles* molar mass

         = 219 mol * 18.0148 g per mol

        = 3.94*10^3 g water

So the mass of water removed = 3.94*10^3 g

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