Question

....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.

Answer 1

@cuanchi
@JoannaBlackwelder

Answer 2

@Abhisar @dan815

Answer 3

Do you know how to balance an equation?

Answer 4

I tried to. I don't know if I got the right answer

Answer 5

@Abhisar

Answer 6

Please show me your work.

Answer 7

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

Answer 8

@Abhisar

Answer 9

@@Abhisar

Answer 10

@Luigi0210

Answer 11

@dan815

Answer 12

@wolverine32

Answer 13

Unbalanced: PbS + O2 -> PbO + SO2

Balanced: 2PbS + 3O2 -> 2PbO + 2SO2

Answer 14

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: \(\sf O_2+2H_2\rightarrow2 H_2O\)

Answer 15

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