mass of hydrogen -> 3 g

moles of hydrogen -> 3/2 = 1.5

mass of nitrogen -> 7 g

moles of nitrogen -> 0.25

now

N2 + 3H2 -> 2NH3

1 mole nitrogen -> 2 mole NH3

0.25 mole nitrogen -> 0.5 mole NH3

since there is Nitrogen as limiting reagent

therefore only 0.5 mole NH3 is formed and 0.75 mole of Hydrigen is left

so mass of 0.5 NH3 = (14 + 3)/2 = 17/2 = 8.5 g