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emmaleelooney

  • one year ago

....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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  1. emmaleelooney
    • one year ago
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    @cuanchi @JoannaBlackwelder

  2. emmaleelooney
    • one year ago
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    @Abhisar @dan815

  3. Abhisar
    • one year ago
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    Do you know how to balance an equation?

  4. emmaleelooney
    • one year ago
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    I tried to. I don't know if I got the right answer

  5. emmaleelooney
    • one year ago
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    @Abhisar

  6. Abhisar
    • one year ago
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    Please show me your work.

  7. emmaleelooney
    • one year ago
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    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

  8. emmaleelooney
    • one year ago
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    @Abhisar

  9. emmaleelooney
    • one year ago
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    @@Abhisar

  10. emmaleelooney
    • one year ago
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    @Luigi0210

  11. emmaleelooney
    • one year ago
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    @dan815

  12. emmaleelooney
    • one year ago
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    @wolverine32

  13. anonymous
    • one year ago
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    Unbalanced: PbS + O2 -> PbO + SO2 Balanced: 2PbS + 3O2 -> 2PbO + 2SO2

  14. aaronq
    • one year ago
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    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: \(\sf O_2+2H_2\rightarrow2 H_2O\)

  15. cuanchi
    • one year ago
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    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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