If 4.11 grams of iodine are reacted with 3.71 grams of H2S, the products are HI and sulfur. I2 + H2S --> 2HI + S What is the theoretical yield of HI?

Answer 1

First, determine the limiting reactant by dividing each mass by its molar mass: I₂ wins at 0.0162 mol, so most of the 0.109 mole of H₂S is not consumed.

The stoichiometry is (4.11g I₂) * (1 mol I₂ / 253.8g I₂) * (2 mol HI / 1 mol I₂) * (127.9g HI / 1 mol HI) = 4.14g HI = theoretical yield.