Answer to Question #178572 in Chemistry for omar

Question

How many gram does nitrogen (N) present in 100 Kg ammonium nitrate ((NH4NO3) fertilizer?

Answer 1

M (NH₄NO₃) = 80.04 g/mol

n (NH₄NO₃) = m / M = 100 000 / 80.04 = 1249.33 mol

The ratio of moles of N to moles of NH₄NO₃ is 2:1. Therefore, N moles is:

n (N) = 1249.33 / 1 x 2 = 2498.66 mol

M (N) = 14.01 g/mol

m (N) = n x M = 2498.66 x 14.01 = 35006.18 g = 35 kg