Question

Ammonium nitrate is one of the most important nitrogenous fertilizers. Its purity can be determined by reacting this ammonium nitrate with Ba(OH)2. During an experiment 0.4082 g fertilizer powder was just converted by adding 24.42 ml 0.1023 M Ba(OH)2 solution. How much % ammonium nitrate is in this fertilizer?

Answer 1

The balanced reaction is : Ba(OH)₂ + 2 NH₄NO₃ Ba(NO₃)₂ + 2 NH₃ + 2 H₂O

Given : concentration of Ba(OH)₂ = 0.1023 M

volume of Ba(OH)₂ consumed = 24.42 mL = 0.02442 L

moles of Ba(OH)₂ consumed = (concentration of Ba(OH)₂) * (volume of Ba(OH)₂ consumed in Liter)

moles of Ba(OH)₂ consumed = (0.1023 M) * (0.02442 L)

moles of Ba(OH)₂ consumed = 0.00250 mol

moles NH₄NO₃ present = (moles of Ba(OH)₂ consumed) * (2 moles NH₄NO₃ / 1 mole Ba(OH)₂)

moles NH₄NO₃ present = (0.00250 mol) * (2 / 1)

moles NH₄NO₃ present = (0.00250 mol) * 2

moles NH₄NO₃ present = 0.00500 mol

mass NH₄NO₃ present = (moles NH₄NO₃ present) * (molar mass NH₄NO₃)

mass NH₄NO₃ present = (0.00500 mol) * (80.04 g/mol)

mass NH₄NO₃ present = 0.400 g

mass % NH₄NO₃ in fertilizer = (mass NH₄NO₃ present / mass of fertilizer) * 100

mass % NH₄NO₃ in fertilizer = (0.400 g / 0.4082 g) * 100

mass % NH₄NO₃ in fertilizer = (0.9797) * 100

mass % NH₄NO₃ in fertilizer = 97.97 %