Determine the limiting reagent for the reaction 2 H2 + O2 → 2 H2O when 3 moles of H2 and 2 moles of O2 are mixed, and state how many moles of H2O can be formed.

Determine the limiting reagent for the reaction 2 H2 + O2 → 2 H2O when 3 moles of H2 and 2 moles of O2 are mixed, and state how many moles of H2O can be formed.

• A. Limiting reagent: O2; moles H2O formed = 3
• B. Limiting reagent: H2; moles H2O formed = 3
• C. Limiting reagent: H2; moles H2O formed = 2
• D. Limiting reagent: O2; moles H2O formed = 2

Answer: (A)

The concept being tested is figuring out which reactant runs out first and using the balanced equation to predict how much product forms. The reaction shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. To use all of the 3 moles of hydrogen, you would need 1.5 moles of oxygen (because 3 H2 × (1 O2 / 2 H2) = 1.5 O2). You actually have 2 moles of O2, which is more than enough, so hydrogen gas runs out first. That makes hydrogen the limiting reagent. With 3 moles of H2 reacting, you form 3 moles of H2O, since the reaction yields 2 H2O for every 2 H2, i.e., H2O formed = H2 consumed = 3 moles. The option that assigns oxygen as limiting would require less oxygen than you have, which isn’t the case here.

The concept being tested is figuring out which reactant runs out first and using the balanced equation to predict how much product forms. The reaction shows that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. To use all of the 3 moles of hydrogen, you would need 1.5 moles of oxygen (because 3 H2 × (1 O2 / 2 H2) = 1.5 O2). You actually have 2 moles of O2, which is more than enough, so hydrogen gas runs out first. That makes hydrogen the limiting reagent. With 3 moles of H2 reacting, you form 3 moles of H2O, since the reaction yields 2 H2O for every 2 H2, i.e., H2O formed = H2 consumed = 3 moles. The option that assigns oxygen as limiting would require less oxygen than you have, which isn’t the case here.