# Daily output of NO from power plant.

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

Last edited:

## Answers and Replies

- 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.

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

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.

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

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, wich weigh ... g, so you get .......... kg of NO

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.

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 :)

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.