Question

question 38 (1 point) listen solid lead(ii)oxide breaks up to give lead metal and gaseous oxygen. the correct balanced equation for this reaction is: 1) $\ce{pb_{3}o_{2}(s) -> pb(s) + o_{2}(g)}$ 2) $\ce{2pb_{2}o(s) -> 2pb(s) + o(g)}$ 3) $\ce{2pbo(s) -> 2pb(s) + o_{2}(g)}$ 4) $\ce{pbo_{2}(s) -> pb(s) + o_{2}(g)}$

Explanation:

Step1: Identify Reactant and Products

The reactant is solid lead(II) oxide, which has the formula \( \text{PbO} \) (since lead(II) has a +2 charge and oxide has -2, so \( \text{Pb}^{2+} \) and \( \text{O}^{2-} \) combine as \( \text{PbO} \)). Products are lead metal (\( \text{Pb}(s) \)) and gaseous oxygen (\( \text{O}_2(g) \)).

Step2: Check Each Option for Correct Formula and Balancing

• Option 1: Reactant is \( \text{Pb}_3\text{O}_2 \), not lead(II) oxide. Incorrect.
• Option 2: Reactant is \( \text{Pb}_2\text{O} \), not lead(II) oxide, and oxygen is \( \text{O}(g) \) (oxygen is diatomic, should be \( \text{O}_2 \)). Incorrect.
• Option 3: Reactant \( 2\text{PbO}(s) \) (correct for lead(II) oxide, 2 moles). Products: \( 2\text{Pb}(s) \) (2 Pb atoms) and \( \text{O}_2(g) \) (2 O atoms, since \( 2\text{PbO} \) has 2 O, which forms 1 \( \text{O}_2 \)). Balanced.
• Option 4: Reactant is \( \text{PbO}_2 \) (lead(IV) oxide, not lead(II) oxide). Incorrect.

Answer:

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

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