Answer
$2.58\times10^4kg$ $NH_{3}$
Work Step by Step
$2NH_{3}+H_{2}SO_{4}$ -> $(NH_{4})_{2}SO_{4}$
mass of $NH_{3}$ (in Kg) = $\frac{(1.00\times10^8g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(2molNH_{3})\times(17.04gNH_{3})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molNH_{3})\times(1\times10^3Kg)}$
=$2.58\times10^4kg$ $NH_{3}$
