\(H_2SO_4 (aq) + 2NH_3(g) \longrightarrow (NH_4)_2 SO_4 (aq)\)

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 H_{2}SO_{4} react with 2 mole of NH_{3} to form = 1 mole (NH_{4})_{2} SO_{4}

\(\therefore \) Number of moles of NH3 required to complete reaction with 20.4 mole of H_{2}SO_{4} = 20.4 x 2 = 40.8 mole

but we have NH_{3} = 58.8 mole

\(\therefore \) H_{2}SO_{4} is a limiting reagent.

\(\therefore \) Number of moles of (NH_{4})_{2} SO_{4} formed = 20.4 mol

\(\therefore \) Mass of (NH_{4})_{2} SO_{4} formed = 20.4 x 132

= 2692.8 g

= 2.7 kg

Hence, maximum mass of (NH_{4})_{2} SO_{4 }can be formed = 2.7 kg.