How much ammonia is required to produce 1 kg of standard ammonium nitrate containing 96% ammonium nitrate?
NH3 + HNO3 = NH4NO3
First, let’s calculate the mass of pure ammonium nitrate
m (NH4NO3) = m (total) * w (NH4NO3) = 1000 * 0.96 = 960 g.
where, 1kg = 1000 gr., and 0.96 – 96% is the content of nitrate
Under the conditions of the problem, it is not said under what conditions the reaction occurs, so we will take them as normal conditions.
Under normal conditions, 1 mol of any gas occupies a molar volume of 22.4 l / mol. It can be seen from the reaction equation that from 1 mol of NH3, 1 mol of NH4NO3 is obtained.
Let’s make the proportion:
X / 22.4 = 960/80
Where M (NH4NO3) = 14 + 4 * 1 + 14 + 3 * 16 = 14 + 4 + 14 + 48 = 80 g / mol
a X is the required volume of ammonia
whence X = (22.4 * 80) / 960 = 1.87 liters.
Answer: the volume of ammonia is 1.87 liters.
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