You combine 10 grams of hydrogen gas (H₂, MW = 2 g/mol) and 10 grams of oxygen gas (O₂, MW = 32 g/mol) for the reaction 2H₂ + O₂ → 2H₂O. Which is the limiting reagent?
AHydrogen, because it has a smaller molecular weight and reacts in a 2:1 ratio with oxygen
BOxygen, because despite equal masses, it provides far fewer moles and the mole-to-coefficient ratio is much smaller
CNeither — equal masses of reactants means they are mixed in stoichiometric proportions
DHydrogen, because it is consumed in a 2:1 ratio and you always need more of the reactant with the larger coefficient
Convert to moles first: H₂ = 10 g ÷ 2 g/mol = 5 mol; O₂ = 10 g ÷ 32 g/mol = 0.3125 mol. Divide by stoichiometric coefficient: H₂ = 5/2 = 2.5; O₂ = 0.3125/1 = 0.3125. The smaller ratio identifies the limiting reagent — O₂ at 0.3125. Despite equal masses, oxygen runs out far sooner: the 5 mol of H₂ would need only 2.5 mol O₂, but only 0.3125 mol O₂ is present. This illustrates why mass comparison is meaningless — you must compare mole-to-coefficient ratios.
Question 2 Multiple Choice
For the reaction N₂ + 3H₂ → 2NH₃, you have 2 mol N₂ and 3 mol H₂. What is the theoretical yield of NH₃?
A4 mol NH₃ — based on 2 mol N₂ reacting fully
B2 mol NH₃ — based on 3 mol H₂ as the limiting reagent
C6 mol NH₃ — based on total moles of reactant
D3 mol NH₃ — based on the average of what each reactant could produce
First identify the limiting reagent: divide each by its stoichiometric coefficient — N₂: 2/1 = 2; H₂: 3/3 = 1. H₂ has the smaller ratio, so H₂ is limiting. Calculate yield from the limiting reagent: 3 mol H₂ × (2 mol NH₃ / 3 mol H₂) = 2 mol NH₃. Theoretical yield is always calculated from the limiting reagent's moles. Using 2 mol N₂ would give 4 mol NH₃, but that assumes N₂ is limiting — it is not; H₂ runs out first, preventing more than 2 mol NH₃ from forming.
Question 3 True / False
If you have more grams of Reactant A than Reactant B in a reaction mixture, Reactant B should be the limiting reagent.
TTrue
FFalse
Answer: False
Grams alone cannot determine the limiting reagent. A large mass of a high-molecular-weight substance may represent far fewer moles than a small mass of a light substance. For example, 100 g of iron (MW = 56) provides 1.79 mol, while 10 g of hydrogen (MW = 2) provides 5 mol. You must always convert to moles and compare mole-to-coefficient ratios. The limiting reagent has the smallest ratio of (moles available)/(stoichiometric coefficient) — which cannot be determined from mass alone.
Question 4 True / False
The theoretical yield is calculated using the stoichiometry of the limiting reagent, because that is the reactant that determines the maximum amount of product that can form.
TTrue
FFalse
Answer: True
Theoretical yield is the maximum amount of product assuming the limiting reagent reacts completely and with 100% efficiency. Once the limiting reagent is consumed, the reaction stops regardless of how much excess reagent remains. Using stoichiometry from the limiting reagent's mole amount gives the ceiling on product formation. Using the excess reagent's moles would overestimate yield — the reaction cannot produce more product than the limiting reagent allows, since excess reagent has leftover material that never reacts.
Question 5 Short Answer
Why must you compare mole-to-coefficient ratios rather than simply comparing masses to identify the limiting reagent?
Think about your answer, then reveal below.
Model answer: The balanced chemical equation specifies the ratios in which substances react in *moles*, not grams — because moles count actual numbers of molecules, and molecules of different substances have very different masses. Comparing masses directly ignores this: 32 grams of oxygen and 2 grams of hydrogen contain the same number of moles (1 mol each), but the equation 2H₂ + O₂ → 2H₂O requires a 2:1 mole ratio of H₂ to O₂. To find which reactant runs out first, divide each reactant's available moles by its stoichiometric coefficient — this tells you how many 'complete recipe sets' each reactant can provide. The one with the fewest sets is the limiting reagent.
The sandwich analogy captures this: if each sandwich requires 2 slices of bread and 1 slice of cheese, dividing by the recipe coefficient (bread: n/2, cheese: n/1) immediately reveals which ingredient limits production. The same logic applies to reactions, which have their own 'recipe' specified by the balanced equation. Always convert to moles first — the equation speaks in moles, and so must you.