Question

Gaseous ammonia can be injected into the exhaust stream of a coal-burning power plant to reduce the pollutant NONO to N2N2 according to the following reaction: 4NH3(g)+4NO(g)+O2(g)→4N2(g)+6H2O(g)4NH3(g)+4NO(g)+O2(g)→4N2(g)+6H2O(g) Suppose that the exhaust stream of a power plant has a flow rate of 335 L/sL/s at a temperature of 955 KK, and that the exhaust contains a partial pressure of NONO of 23.2 torrtorr .

Part A What should be the flow rate of ammonia delivered at 765 torrtorr and 298 KK into the stream to react completely with the NONO if the ammonia is 65.0% pure (by volume)?

Answer 1

Step 1

Given that the pressure of NO, PNO = 23.2 torr.

Convert into atm where 1atm = 760torr, so the 23.2 torr(1atm/760torr) => 0.0305 atm

Power plant volume flowrate () NO = 335 L/s, temperature of NO(TNO) = 955 K.

By ideal gas equation, PV=nRT

For NO apply ideal gas equation,

PV = nRT

Divide both side by t we get,

P(V/t) = (n/t)RT

(n/t)NO = 0.1303 mol/s

Step 2

convert (n/t) ratio from NO to NH3 using the mole ratio based on the stoichiometric coefficient of the chemical equation.

the stoichiometric coefficient for NO and NH3 is the same(4) so we can write,

On cancellation of mol of NO we have,

(n/t)NH3 = 0.1303 mol NH3

Now find out the pressure of ammonia in atm which is given as 765 torr,

PNH3 = 765torr(1atm/760torr) => 1.01 atm

The temperature of NH3 = 298 K.

By applying ideal gas equation on NH3 we have, P(V/t) = (n/t)RT.

Find out the (V/t) for pure NH3,

(V/t)NH3 = 3.155 L/s

Now change the percentage into a decimal and divide quantity to get the ((V/t)NH3 for pure ammonia.

((V/t)NH3)/0.65 = (3.155L/s)/0.65 => 4.85 L/s Flow rate of ammonia at 765 torr.