Stoichiometry and limiting reagent
Mole Concept and Stoichiometry · Chemistry
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4 g H₂ reacts with 32 g O₂: 2H₂ + O₂ → 2H₂O. Find limiting reagent and water formed. Step 1: Moles: H₂ = 4/2 = 2 mol; O₂ = 32/32 = 1 mol. Step 2: Required ratio H₂:O₂ = 2:1. Available = 2:1 — exact match! Step 3: Neither is limiting here (both consumed fully). Step 4: Water = 2 mol × 18 = 36 g. If only 1 mol H₂ were present: H₂ would limit, giving 1 mol H₂O = 18 g.