\(H_2SO_4 (aq) + 2NH_3(g) \longrightarrow (NH_4)_2 SO_4 (aq)\)
Number of moles of \(H_2SO_4 = \frac{2000}{98} = 20.4 \,moles \)
Number of moles of \(NH_3 = \frac{1000}{17} = 58.8\, moles\)
\(\because\) 1 mole of H2SO4 react with 2 mole of NH3 to form = 1 mole (NH4)2 SO4
\(\therefore \) Number of moles of NH3 required to complete reaction with 20.4 mole of H2SO4 = 20.4 x 2 = 40.8 mole
but we have NH3 = 58.8 mole
\(\therefore \) H2SO4 is a limiting reagent.
\(\therefore \) Number of moles of (NH4)2 SO4 formed = 20.4 mol
\(\therefore \) Mass of (NH4)2 SO4 formed = 20.4 x 132
= 2692.8 g
= 2.7 kg
Hence, maximum mass of (NH4)2 SO4 can be formed = 2.7 kg.