How many liters of ammonia and what mass of a 68% sulfuric acid solution will be needed

| September 23, 2021 | education

How many liters of ammonia and what mass of a 68% sulfuric acid solution will be needed to obtain 1 ton of ammonium sulfate if the reaction product yield is 60%?

Given: Solution:
m ((NH4) 2SO4 = 1 t Find 100% of the product: 100% – x
w (H2SO4) = 68% 60% – 1t
Product yield 60% x = 1 * 100/60 = 1.67 t = 1670 kg = 1670000 g.
Find: 2NH3 + H2SO4 = (NH4) 2SO4
V (NH3) =? x g – 1670000 g
m (p-pa) =? 98 – 132 g / mol
x = 98 * 1670000/132 = 1239848.5 g-mass of sulfuric acid
Determine the mass of the sulfuric acid solution:
m (p-pa) = m (H2SO4) * 100 / M = 1239848.5 * 100/68 =
= 1823306.6 g = 1823.3 kg
Determine the mass of ammonia:
x g – 1670000 g
17 – 132 g / mol
x = 17 * 1670000/132 = 215075.7 g mass of ammonia.
Ammonia volume: V (NH3) = m * 22.4 / M =
215075.7 * 22.4 / 17 = 283393.94 = 283394 hp
Answer: m (p-pa) = 1823.3 kg, V (NH3) = 283394 l



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