What is the limiting reagent when 49.84 g of nitrogen react with 10.7 g of hydrogen according to this balanced equation: n2 (g) 3h2 (g) → 2nh3 (g) ?

Moles of nitrogen = 49.84 / 28 = 1.78

Moles of hydrogen = 10.7 / 2 = 5.35

Molar ratio of nitrogen to hydrogen = 1 : 3

Hydrogen required = 1.78 x 3 = 5.34

Therefore, hydrogen is in excess and nitrogen is the limiting reagent.