Answer
20.96g $H_{2}SO_{4}$
Work Step by Step
a. $2NH_{3}(aq)+1H_{2}SO_{4}(aq)$ -> $ 1(NH_{4})_{2}SO_{4}(aq)$
b.
Mass of $NH_{3}$ = $\frac{(20.3g(NH_{4})_{2}SO_{4})\times((NH_{4})_{2}SO_{4})\times(2molNH_{3})\times(17.04molNH_{3})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molNH_{3})}$
= 5.23 g $NH_{3}$
Mass of $H_{2}SO_{4}$= $\frac{(20.3g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molH_{2}SO_{4})\times(98.09gH_{2}SO_{4})}{(132.17g(NH_{4})_{2}SO_{4})\times(1mol(NH_{4})_{2}SO_{4})\times(1molH_{2}SO_{4})}$
= 15.07 g
15.07 + 5.89 = 20.96g $H_{2}SO_{4}$
