Question

if 5.1 g of h₂ were to be mixed with 20.52 g of o₂, 2 h₂ + o₂ → 2 h₂o, the limiting reagent would be 1. o₂. 2. h₂o. 3. the reactants are present in stoichiometric amounts. 4. h₂.

Explanation:

Step 1: Calculate moles of \( H_2 \)

Molar mass of \( H_2 = 2 \, g/mol \). Moles of \( H_2 = \frac{5.1 \, g}{2 \, g/mol} = 2.55 \, mol \).

Step 2: Calculate moles of \( O_2 \)

Molar mass of \( O_2 = 32 \, g/mol \). Moles of \( O_2 = \frac{20.52 \, g}{32 \, g/mol} = 0.64125 \, mol \).

Step 3: Check stoichiometric ratio

From reaction \( 2H_2 + O_2
ightarrow 2H_2O \), ratio \( H_2:O_2 = 2:1 \).
For \( O_2 = 0.64125 \, mol \), required \( H_2 = 2 \times 0.64125 = 1.2825 \, mol \).
Available \( H_2 = 2.55 \, mol \) (more than required). So \( O_2 \) is limiting? Wait, no—wait, let's recheck. Wait, maybe I mixed up. Wait, let's do it the other way: for \( H_2 = 2.55 \, mol \), required \( O_2 = \frac{2.55}{2} = 1.275 \, mol \). But available \( O_2 = 0.64125 \, mol \) (less than 1.275). Wait, no, wait my first calculation was wrong. Wait, moles of \( O_2 \): \( 20.52 / 32 = 0.64125 \, mol \). Moles of \( H_2 \): 5.1 / 2 = 2.55 mol. The reaction ratio is 2 H₂ to 1 O₂. So for H₂: 2.55 mol, the O₂ needed is 2.55 / 2 = 1.275 mol. But we only have 0.64125 mol O₂, which is less. Wait, but that would mean O₂ is limiting? But the options have O₂ as option 1. Wait, no, maybe I made a mistake. Wait, wait the question's options: let's re-express. Wait, maybe I flipped. Wait, let's check the required moles again. Wait, the reaction is 2 H₂ + 1 O₂. So moles of H₂: 2.55, moles of O₂: 0.64125. The ratio of H₂ to O₂ is 2.55 / 0.64125 ≈ 3.976, which is more than 2 (the stoichiometric ratio 2:1). So O₂ is the limiting reagent? Wait, but the options: 1. O₂, 4. H₂. Wait, no, wait maybe I messed up the calculation. Wait, 5.1 g H₂: moles = 5.1 / 2 = 2.55 mol. 20.52 g O₂: 20.52 / 32 = 0.64125 mol. The stoichiometric ratio is 2 H₂ per 1 O₂. So for O₂ = 0.64125 mol, H₂ needed is 2 0.64125 = 1.2825 mol. We have 2.55 mol H₂, which is more than 1.2825. So O₂ is limiting? But the options have O₂ as option 1. Wait, but let's check again. Wait, maybe the question is different. Wait, maybe I miscalculated O₂ moles. 20.52 divided by 32: 320.6 = 19.2, 20.52 - 19.2 = 1.32, 1.32 /32 = 0.04125, so total 0.64125 mol. Correct. H₂: 5.1 /2 = 2.55 mol. So the O₂ is insufficient to react with all H₂. So O₂ is the limiting reagent. So the answer is option 1: O₂.

Answer:

1. \( \text{O}_2 \)