Reacting 3.0 mol of H2 with 2.0 mol of I2 gave 1.0 mol of HI. What is the percent yield for H2 + I2 giving 2HI?

1. The equation needs hydrogen and iodine in a 1 to 1 mole ratio. 2. With 3.0 mol H2 and 2.0 mol I2, iodine is the limiting reactant. 3. Complete reaction of 2.0 mol I2 would give 2 times 2.0, that is 4.0 mol HI. 4. Hydrogen alone could have given 6.0 mol HI, confirming that iodine caps the yield. 5. So the theoretical yield is 4.0 mol HI and the actual yield is 1.0 mol HI. 6. Percent yield is 1.0 divided by 4.0, times 100, which is 25%. 7. Using 3.0 mol H2 as the basis would wrongly give about 17%, which is the limiting-reactant trap. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.4 Reaction Yields_