....hmmmm kinda question... Please help me! The following unbalanced equation describes the reaction that can occur when lead (lll) sulfide reacts with oxygen gas to produce lead (lll) oxide and sulfur dioxide gas: PbS + O2 ----> PbO +SO2 Balance the equation and describe in words the electron transfer(s) that takes place.

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....hmmmm kinda question... Please help me! The following unbalanced equation describes the reaction that can occur when lead (lll) sulfide reacts with oxygen gas to produce lead (lll) oxide and sulfur dioxide gas: PbS + O2 ----> PbO +SO2 Balance the equation and describe in words the electron transfer(s) that takes place.

Chemistry
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Do you know how to balance an equation?

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I tried to. I don't know if I got the right answer
Please show me your work.
I subtracted O2 on each side, canceled it out.. So that gave me PbS------>PbO+S PbO+S+S+O. Which I ended up with Pb2+S2+O
Unbalanced: PbS + O2 -> PbO + SO2 Balanced: 2PbS + 3O2 -> 2PbO + 2SO2
When you balance reactions, youre only allowed to change the coefficients, even though there might be no numbers explicitly written, there is a 1. \(\sf O_2+H_2\rightarrow H_2O\) means: \(\sf \color{red}1~O_2+\color{red}1~H_2\rightarrow \color{red}1~H_2O\) when balanced: \(\sf \color{red}1~O_2+\color{red}2~H_2\rightarrow \color{red}2~H_2O\) written as: \(\sf O_2+2H_2\rightarrow2 H_2O\)
In your question mention "lead (lll) sulfide reacts with oxygen gas to produce lead (lll) oxide" The roman numeral, III, indicates that in this case lead would have an oxidation state of 3+. When this is in a compound with the sulfide anion, S2-, the formula would be Pb2S3. HOWEVER, lead is not really found with a 3+ oxidation state- only 0, 2+, or 4+. Lead (II) sulfide= PbS Lead (IV) sulfide= PbS2

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