The question is incomplete, the complete question is;
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?
Answer:
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
