Wetlands can play a significant role in removing fertilizer residues from rain runoff and groundwater; one way they do this is through denitrification, which converts nitrate ions to nitrogen gas:

Wetland is a natural region of marches, pen and peatlands of water. The wetlands have static water and this forms a distant ecosystem that may be flooded by water permanently or seasonally.

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Wetlands can play a significant role in removing fertilizer residues from rain runoff and groundwater; one way they do this is through denitrification, which converts nitrate ions to nitrogen gas:

2NO_3^-(aq) + 5 CO(g) + 2 H+(aq)  N2(g) + H2O(l) + 5 CO2(g)

Supposed 200.0 g of NO_3^- flows into a swamp each day. What volume of N2 would be produced at 17 deg C. and 1.00 atm if the denitrification process were complete? What volume of CO2 would be produced? Suppose the gas mixture produced by the decomposition reaction is trapped in a container at 17 deg C; what is the density of the mixture if we assume that the total P = 1.00 atm?  

See detailed solution below

Explanation:

First we must obtain the number of moles of NO3^- from

number of moles = mass/molar mass

mass of = 200g

molar mass of NO3^- = 62 g/mol

number of moles = 200g/62g/mol = 3.23 moles

From the reaction equation;

2 moles of NO3^- yields 1 mole of N2

3.23 moles yields 3.23 * 1/2 = 1.6 moles

From

PV = nRT

V = nRT/P

V = 1.6 * 0.082 * 290/1

V= 38 L

Then;

From the equation also;

2 moles of  NO3^-  yields 5 moles of CO2

3.23 moles of NO3^-  yields 3.23 * 5/2 = 8.1 moles

So,

PV=nRT

V = nRT/P

V= 8.1 * 0.082 * 290/1

V= 192.6 L

Lastly

The temperature of the mixture is 17 degrees Celsius and the pressure of the mixture is 1 atm

the total number of moles of both gases = 1.6 + 8.1 = 9.7 moles

Volume of the mixture = nRT/P = 9.7 *0.082*290/1 = 230.67 L

Mass of N2 = 1.6 moles * 28 g/mol = 44.8 g

Mass of CO2 = 8.1 moles * 44 g/mol = 356.4 g

Total mass of gases = 44.8 + 356.4 = 401.2 g

Density of mixture = mass/volume = 401.2g/230.67 L

Density of mixture = 1.74 g/L