Question

A golf course manager applies 500 lbs of ammonium nitrate per acre each year . Every four years he applies CaCO 3 to counteract the effects of the ammonium nitrate applied during the previous four years . How much CaCO 3 (100% fineness) would need to be applied every four years (in lbs/acre ) to offset the NH 4 NO 3 effects on the pH of the soil? Ammonium nitrate c ontains 34% N and undergoes the following reactions in soil.

NH4NO3 -->NH4+ + NO3-

NH4+ + 2 O2 --> NO3- + H2O + 2H+

Answer 1

m = 500 lb NH4NO3/(acre) per year

mass (g) of NH4NO3 = mass * 454 g/lb = 500*454 = 227000 g

mol of NH4NO3 = mass/MW = 227000/80.0434  = 2835.961 mol of NH4NO3

note that

1 mol of NH4NO = 1 mol of NH4+

1 mol of NH4+ = 2 mol of H+

then

2835.961mol of NH4 = 2*2835.961 = 5671.922 mol of H+

note that

CaCO3 + 2H+ --> Ca+2 + H2O + CO2

then,

5671.922 mol of H+ --> 1/2*5671.922 = 2835.961 mol of CaCO3 required

mass = mol*MW = 2835.961*100 = 283596.1 g of CaCO3 required per 4 year period

change to lb --> 283596.1 g --> 283596.1/454 = 624.661 lb of CaCO3 per acre