In a lab you react 4.3 mols of hydrogen gas with 2.3 mols of oxygen gas to form water. You collect 12g of water. What is the percentage yield?

Answer 1

Using mole ratios from balanced reaction 2 H₂ + O₂ --> 2 H₂O:

From hydrogen gas, (4.3 mol H₂) * (2 mol H₂O /2 mol H₂) = 4.3 mole H₂O.

From oxygen gas, (2.3 mol O₂) * (2 mol H₂O /1 mol O₂) = 4.6 mole H₂O. Hydrogen is limiting.

Then (4.3 mol H₂O) * (18.0g H₂O /1 mol H₂O) = 77g H₂O theoretical yield.

Percent yield = 100 actual/theoretical = 100 * 12/77 = 16%.