Question

The fertilizer ammonium sulfate is prepared by the reaction between ammonia and sulfuric acid.How many kg of ammonia are needed to produce 1x10^5 kg of (NH4)2SO4?

Answer 1

\[NH_3 + H_2SO_4 -> (NH_4)_2SO_4\]
Molar mass of (NH4)2SO4 = (14.0 + 1.0x4)x2 + 32.1 + 16.0x4 = 132.1
No. of mole of (NH4)2SO4 =mass/ molar mass = 1x10^5 x 1000 / 132.1 = 757002.271 mole
No. of mole of ammonia needed = No. of mole of (NH4)2SO4 = 757002.271 mole
Mass of ammonia needed = 757002.271 x mole mass of NH3
= 757002.271 x (14.0 + 1.0x3)
= 757002.271x 17.0
= 12869038.61 g
= 1.29 x10^4 kg