Daily output of NO from power plant.

[eddzzz_2011]

Homework Statement :

Q.If a power plant burning 10,000 metric tons of coal per day with 10% excess air emits stack gas containing 100 ppm by volume of NO, what is the daily output of NO? (Assume coal is pure C).


The attempt at a solution:

A. C + O2 → CO2

(10000 metric tons of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 1000000) = 3.66 metric tons of NO

Please can someone check this is correct for me. If incorrect please show me. Thanks,

Ed

 

[willem2]

- You forgot the 10% excess air
- the atmosphere is only 21% oxigen (by volume)
- you calculated 100 ppm by weight and not by volume

start with calculating how many moles of C are present. You can then compute
how many moles of NO are present without using the weight of anything, and then
finally compute the weight of the NO.

 

[eddzzz_2011]

Moles of C present:

10,000 / 12.0108 = 832.58 moles of coal. (do i need to convert 10,000 metric tons into grams? Other than that I have no clue what to do with the 100ppm by the volume. Please show me the working out to receive the correct answer then I can go through it and understand better. Thanks

 

[willem2]

Moles of C present:

10,000 / 12.0108 = 832.58 moles of coal. (do i need to convert 10,000 metric tons into grams?

Yes, 1 mole of carbon is 12.0108 g,and not 12.0108 metric tons.

Other than that I have no clue what to do with the 100ppm by the volume. Please show me the working out to receive the correct answer then I can go through it and understand better. Thanks

the volume of 1 mole is the same for all the gases, so if 100 ppm by volume is NO, there's one 1 mole of NO, for every 10000 moles of gas coming out of the pipe.

you need to add a volume of air that contains enough oxygen to react with the coal. Air is 21% oxygen by volume, so you need 100 moles of air to get 21 moles of oxygen and then you need 10% extra air.

 

[eddzzz_2011]

10,000 metric tons = 10 000 000 000 grams. (1.0 * 10^10)

1 mole of C = 832.58 moles
1 mole of NO = 832.58 moles

So the equation would be:

(1.0*10^10 grams of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 10000) = 366416891.5 grams of NO?? Sorry I really don't understand this one

 

[willem2]

10,000 metric tons = 10 000 000 000 grams. (1.0 * 10^10)

1 mole of C = 832.58 moles
1 mole of NO = 832.58 moles

So the equation would be:

(1.0*10^10 grams of C) / 12.0108 g/mol x (1/1) x (44.0096 g/mol) x (100/ 10000) = 366416891.5 grams of NO?? Sorry I really don't understand this one

don't try to write everyting in one equation, that way I don't understand it either.

there are 10^10 / 12.0108 = 8.33 *10^8 moles of C present.

these need 8.33 *10^8 moles of oxygen to burn

1 mole of air contains 0.21 moles of oxygen so

... moles of air contain 8.33 *10^8 moles of oxygen

there's 10% extra air, so you need .... moles

100/1000000 parts of the exhaust is NO, so this is ... moles, which weigh ... g, so you get ... kg of NO

 

[eddzzz_2011]

Yeah shouldn't of put it in that way. Ill put it in a clearer way in future.

1 mole of air contains 0.21 moles of oxygen so

3.9*10^9 moles of air contain 8.33 *10^8 moles of oxygen

there's 10% extra air, so you need 3.57*10^9 moles

100/1000000 parts of the exhaust is NO, so this is 3.57*10^5 moles,

which weigh 10713570g, so you get 10713.570 kg of NO.

 

[eddzzz_2011]

Ignore post above (wrong)

1 mole of air contains 0.21 moles of oxygen so

3.9*10^9 moles of air contain 8.19*10^8 moles of oxygen

there's 10% extra air, so you need 1.209*10^9 moles

100/1000000 parts of the exhaust is NO, so this is 1.209*10^5 moles,

which weigh 3.638*10^6g, so you get 3628kg of NO. Willem2 please check :)

 

[eddzzz_2011]

Assume air is 21% O2 and is at STP (273 K, 1.00 atm)

From the stoichiometry of the reaction C + 02 --> CO2 the mass of 02 needed is given by:

Mass O2 = 10,000 tonne x (32 tonne O2/ 12 tonne C) 2.67 x 10^4 tonne O2

Stoichiometric mass air 2.67 x 10^4 tonne/0.21 = 1.27 x 10^5 tonne air 1.27 x 10^11 g air, To account for 10% excess air:

Mass air = 1.27 x 10^5 tonne x 1.10 = 1.40 x 10^5 tonne air 1.40 x 10^11 g air

Using 1.29 g/L for the density of air, its volume is given by

V 1.40 X 10^11 g/1.29 g/L = 1.09 x 10^11 L

Mol air = 1.09 x 10^11 L/(22.4 L/mol) = 4.84 x 10^9 mol air

At 100 ppm NO, the mol fraction of NO is 1.00 x 10^-4

Mol NO = 1.00 x 10^-4 mol NO/mol air x 4.84 x 10^9 mol air = 4.84 x 10^5 mol NO

Mass NO = 4.84 x 10^5 mol NO x 30.0 g NO/mol NO = 1.45 x 10^7 g NO

The daily output of NO is 14.5 tonne/day.

This seems an easier way.

Could anyone check please.