Is the stoichiometric equation balanced? write down the number of atoms of each

Question

Is the stoichiometric equation balanced? write down the number of atoms of each species on left side and also on the right side of this equation. In the absence of side reactions, what is the yield of nitrogen from acetic acid in g g-1? A certain waste water contains 6.0 mM acetic acid and 7mM NaNO3. If 25% of the acetic acid and 15% of the nitrate (NO3) are consumed in other reactions (e.g, for growth of the microorganisms in the sludge), which is the limiting substrate in the denitrification reaction? For the situation described in part c), what mass of gaseous nitrogen (N2) in grams is produced from treatment of 5000 liters of waste water if the reaction is allowed to proceed until the limiting substrate is exhausted? What is the extent of reaction xi under these conditions?

Explanation / Answer

a) Given equation,

5CH3COOH + 8NO3^- ----> 4N2 + 10CO2 + 6H2O + 8OH-

is balanced,

On reactant side         On product side

C = 10                            C = 10

H = 20                            H = 20

N = 8                              N = 8

O = 34                            O = 34

b) In the absence of side reaction,

5 moles of CH3COOH would give 4 moles of N2

So, 60.05 g would give 4 x 60.05/5 = 48.04 g of N2

60.05 g of acetic acid would yield 48.04 g of N2

c) concentration of acetic acid in waste water = 6 mM = 0.006 M

concentration of NO3- = 7 mM = 0.007 M

25% of 0.006 M = 0.0015 M of acetic acid and 15% of 0.007 M = 0.0015 M of NO3- is consumed in other reactions

Remaining acetic acid = 0.0045 M

Remaining NO3- = 0.0055 M

If all of acetic acid is consumed we would need = 0.0045x8/5 = 0.0072 M of NO3-

If all of NO3- is consumed we would need = 0.0055 x 5/8 = 0.00344 M of CH3COOH

Since we only have 0.0055 M of NO3-, this would be the limiting reagent in the reaction. Note, acetic acid is in excess to the required value.

d) For 1 L in part c) the amount of N2 produced would be,

moles of NO3- = 0.0055 x 5000 = 27.5 moles

mass of N2 produced = 4 x 27.5/8 = 13.75 moles x 28 = 385 g

Is all of original NO3- was used we would get,

= 0.007 x 5000 = 35 moles x 4/8 = 17.5 moles x 28 = 490 g of N2

Extent of reaction = 385/490 x 100 = 78.57%

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