Final answer:
The limiting reagent in a chemical reaction can be determined by comparing the stoichiometric amounts of each reactant to their actual amounts. In the reaction 2 H₂(g) + O₂(g) → 2 H₂O(g), O₂ is the limiting reagent when given 5 grams of H₂ and 10 grams of O₂.
Step-by-step explanation:
The limiting reagent in a chemical reaction is the reactant that is completely consumed and limits the amount of product that can be formed. In the reaction 2 H₂(g) + O₂(g) → 2 H₂O(g), the limiting reagent can be determined by comparing the molar ratios of H₂ and O₂ in the balanced equation to the actual amounts of each reactant given. Whichever reactant has the smaller stoichiometric amount relative to the other is the limiting reagent. In this case, we can compare the moles of H₂ and O₂ to see which one is limiting.
To do this, we need to know the molar masses of H₂ and O₂ and convert the given masses of each reactant to moles using the molar mass. Once we have the moles of H₂ and O₂, we can divide by their stoichiometric coefficients from the balanced equation to determine the stoichiometric amounts. From there, we can compare the stoichiometric amounts of H₂ and O₂ to see which one is limiting.
Let's say we have 5.0 grams of H₂ and 10.0 grams of O₂. The molar mass of H₂ is 2 g/mol and the molar mass of O₂ is 32 g/mol. Converting the given masses to moles, we have:
H₂: (5.0 g / 2 g/mol) = 2.5 mol
O₂: (10.0 g / 32 g/mol) = 0.3125 mol
Next, we divide the moles of H₂ and O₂ by their stoichiometric coefficients:
H₂: 2.5 mol / 2 = 1.25 mol
O₂: 0.3125 mol / 1 = 0.3125 mol
From the calculations, we can see that the stoichiometric amount of O₂ is smaller than that of H₂. Therefore, O₂ is the limiting reagent in this reaction.