For the chemical reaction below, determine the amount of hi produced when 3.35 g of hydrogen is reacted with 50.75 g of iodine to produce hydrogen iodide

Molar mass of H₂ = 1.008 * 2 g/mol = 2.016 g/mol

Molar mass of I₂ = 126.9 * 2 g/mol = 253.8 g/mol

Molar mass of HI = (1.008 + 126.9) g/mol = 127.9 g/mol

H₂ (g) + I₂ (g) → 2HI

Mole ratio H₂ : I₂ : HI = 1 : 1 : 2

Then the initial number of moles of H₂ = (3.35 g) / (2.016 g/mol) = 1.662 mol

Initial number of moles of I₂ = (50.75 g) / (253.8 g/mol) = 0.2000 mol < 1.662 mol

Hence, I₂ is the limiting reactant (limiting reagent).

Number of moles of I₂ reacted = 0.2000 mol

Number of moles of HI reacted = (0.2000 mol) * 2 = 0.4000 mol

Mass of HI reacted = (127.9 g/mol) * (0.4000 mol) = 51.16 g