Question

Large quantities of fertilizer are washed into the Mississippi River from agricultural land in the Midwest. The excess nutrients collect in the Gulf of Mexico, promoting the growth of algae and edging out other aquatic life. Corn farmers typically use 5000 kg of ammonium nitrate per square kilometer of cornfield per year. Ammonium nitrate can be prepared by the following reaction. NH3?+HNO3? -> NH₄NO₃

1. How much nitric acid would be needed to make the fertilizer needed for 1 km2 of cornfield per year? _____ kg nitric acid

2. The ammonium ions can be converted into NO3– by bacterial action. If 10% of the ammonium component of 5000 kg of fertilizer ends up as nitrate, how much oxygen would be consumed? NH^{+ ₄} +2O₂-> NO^-₃+H₂O+2H⁺ _____ kg oxygen

Answer 1

1) Corn farmers typically use 5000 kg of ammonium nitrate per square kilometer of cornfield per year.

So moles of ammonium nitrate required per square kilometer of cornfield per year = Mass / Molar mass = 5000 X 1000g / 80 =62500 moles

as per balanced equation, one mole of nitric acid is required for one mole of ammonium nitrate

so for 62500 moles of ammonium nitrate, we need 62500 moles of nitric acid

mass of nitric acid required = moles X molar mass = 62500 X 63 = 3937500 grams = 3937.500 Kg

2) NH₄⁺+2O₂-> NO^-₃+H₂O+2H⁺

10% of 5000 Kg = 500 Kg

Moles of ammonium nitrate = 500000 / 80= 6250 moles

Moles of ammonium ion = 6250 moles

Moles of oxygen required = 2 X 6250 = 12500 moles

Mass of oxygen required = Moles X molar mass = 4 X 10⁵ grams = 400 Kg