1. Solve the mass of NH4+ ions = (3%/100%)(1x10^4kg) = 300 kg = 300,000 gÂ
2. Compute for the amount that will be converted as a bacterial tissue:Â
NH4+ ions consumed by bacteria = (95%/100%)(300,000 g) = 285,000 gÂ
3. Convert the mass of NH4+ ion calculated in step 2 into the # of moles:Â
moles of NH4+ = mass of NH4+ / MW of NH4Â
moles of NH4+ = 285,000 g / 18 g/mol = 15,833.33 mol
the stoichiometric ratio of C5H7O2N and NH4 is 1 : 55 grounded on the given balanced chemical equation:Â
moles of C5H7O2N = moles of NH4+ x (1 mole of C5H7O2N / 55 moles of NH4+)Â
moles of C5H7O2N = 15,833.33 moles x (1/55) = 303.03 molesÂ
4. Convert the moles of C5H7O2N into grams:Â
mass of C5H7O2N = # of moles of C5H7O2N x MW of C5H7O2NÂ
= 303.03 moles x 113 g/molÂ
= 34,242.39 grams
= 34.34 kgÂ