How many grams of H2 are needed to produce 14.88 g of NH3?

Molar mass of H₂ = (1.008 × 2) g/mol = 2.016 g/mol
Molar mass of NH₃ = (14.01 + 1.008×3) g/mol = 17.034 g/mol

N₂(g) + 3H₂(g) → 2NH₃(g)
OR: Mole ratio H₂ : NH₃ = 3 : 2

No. of moles of NH₃ produced = (14.88 g) / (17.034 g/mol) = 0.8735 mol
No. of moles of H₂ needed = (0.8735 mol) × (3/2) = 1.310 mol
Mass of H₂ needed = (1.310 mol) × (2.016 g/mol) = 2.641 g

Atomic mass of N = 14.01
Atomic mass of H = 1.008
Molar mass of NH3 = 17.03
The 14.88 g of NH3 will contain
(3.024/17.03) * 14.88 g of hydrogen;
that's 2.64 grams,
so that's how many grams of H2 are needed.