Question

1. what is the balanced equation for the formation of rust if four moles of iron metal react with three moles of oxygen gas to produce two moles of iron (iii) oxide. options: 4 fe (s) + 3 o₂ (g) → 2 fe₃o (s); 4 fe (s) + 3 o₂ (g) → 2 fe₂o₃ (s); fe₄ (s) + o₃ (g) → fe₃o₂ (s); fe₄ (s) + o₃ (g) → fe₂o₃ (s); clear all

Explanation:

1. First, recall the formula for iron(III) oxide, which is $\ce{Fe2O3}$ (since iron has a +3 charge and oxygen has a -2 charge, to balance charges: 2 Fe³⁺ and 3 O²⁻ give $\ce{Fe2O3}$).
2. The reactants are iron metal ($\ce{Fe(s)}$) and oxygen gas ($\ce{O2(g)}$). The problem states 4 moles of Fe, 3 moles of $\ce{O2}$, and 2 moles of the product.
3. Check each option:

• Option 1: Product is $\ce{Fe3O}$, which is not iron(III) oxide. Eliminate.
• Option 2: Reactants are 4 $\ce{Fe(s)}$ and 3 $\ce{O2(g)}$, product is 2 $\ce{Fe2O3(s)}$. Let's check atom counts:
• Fe: Left = 4, Right = 2×2 = 4 (balanced).
• O: Left = 3×2 = 6, Right = 2×3 = 6 (balanced). And the product is iron(III) oxide ($\ce{Fe2O3}$).
• Option 3: Reactants are $\ce{Fe4(s)}$ (incorrect formula for iron metal, should be $\ce{Fe}$) and $\ce{O3(g)}$ (oxygen gas is $\ce{O2}$), product is $\ce{Fe3O2}$ (not iron(III) oxide). Eliminate.
• Option 4: Reactants have incorrect formulas ($\ce{Fe4}$ and $\ce{O3}$), and product doesn't match the moles. Eliminate.

Answer:

B. $4 \text{ Fe (s)} + 3 \text{ O}_2 \text{ (g)}
ightarrow 2 \text{ Fe}_2\text{O}_3 \text{ (s)}$