Question

lead(ii)oxide breaks up to give lead metal and gaseous diatomic oxygen. the correct balanced equation for this reaction is

1. $\ce{pb3o2(s) -> 3pb(s) + o2(g)}$
2. $\ce{pbo(s) -> pb(s) + o(g)}$
3. $\ce{2pbo(s) -> 2pb(s) + o2(g)}$
4. $\ce{pbo2(s) -> pb(s) + o2(g)}$

Explanation:

Step1: Identify Reactant and Products

The reactant is Lead(II) oxide, which has the formula \( \text{PbO} \) (since lead is +2 and oxygen is -2, \( \text{Pb}^{2+} \) and \( \text{O}^{2-} \) combine as \( \text{PbO} \)). The products are lead metal (\( \text{Pb}(s) \)) and diatomic oxygen (\( \text{O}_2(g) \)).

Step2: Analyze Each Option

• Option 1: Reactant is \( \text{Pb}_3\text{O}_2 \), not Lead(II) oxide (\( \text{PbO} \)), so incorrect.
• Option 2: Product oxygen is \( \text{O}(g) \), but oxygen is diatomic (\( \text{O}_2 \)) in gas state, so incorrect.
• Option 3: Reactant is \( 2\text{PbO}(s) \), products are \( 2\text{Pb}(s) \) and \( \text{O}_2(g) \). Check balancing:
• Pb: Left = 2, Right = 2 (balanced).
• O: Left = \( 2 \times 1 = 2 \), Right = \( 2 \) (from \( \text{O}_2 \)), balanced.
• Option 4: Reactant is \( \text{PbO}_2 \) (Lead(IV) oxide, since \( \text{Pb}^{4+} \) and \( \text{O}^{2-} \)), not Lead(II) oxide, so incorrect.

Answer:

1. \( 2\text{PbO}(s)

ightarrow 2\text{Pb}(s) + \text{O}_2(g) \)