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balancing

can anyone tell me how i can balance this equation: Al203 = Al (l) + O2(g)
into this balanced equation: 2Al203(l) =4Al(l)+ 3O2(g)
Start with oxygen- the LCM of 3 & 2 is 6; there should be 6 O on each side (or a multiple of 6).

This gives 4 Al on left- must be equal on right

(see attached working out for clarity)
Reply 2
Since there are X times 2 oxygen on one side and Y times 3 on the other and X and Y are both the smallest possible numbers above 0 it is quite simple to deduce that X=3 and Y=2 .
( 2X=3Y )
After writing that down the number of Al isnt equal on both sides so you muliply that getting 4 Al on the right side. My advice if you have problems with that is to raise every number until the total numbers of atoms of all types are equal on both sides by multyplying them by the number on the other side and after you finish doing that just find the biggest number you can divide those with.
Original post by L Addison
Start with oxygen- the LCM of 3 & 2 is 6; there should be 6 O on each side (or a multiple of 6).

This gives 4 Al on left- must be equal on right

(see attached working out for clarity)

thank you so much for taking your time to answer!! this really helped me
Original post by Elize W
Since there are X times 2 oxygen on one side and Y times 3 on the other and X and Y are both the smallest possible numbers above 0 it is quite simple to deduce that X=3 and Y=2 .
( 2X=3Y )
After writing that down the number of Al isnt equal on both sides so you muliply that getting 4 Al on the right side. My advice if you have problems with that is to raise every number until the total numbers of atoms of all types are equal on both sides by multyplying them by the number on the other side and after you finish doing that just find the biggest number you can divide those with.

Thank you so much!!! this really explained it well! thanks for taking your time to answer
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