Antimony is alloyed with lead to increase the rigidity of components used in the construction of lead storage batteries.
1.340 g of a particular metallic alloy, compounded of only Pb and Sb, can be quantitatively converted into a 1.558-g mixture of the oxides PbO2 and Sb2O4.
What was the percentage (by mass) of antimony in the alloy?
Okay so I'm having a bit of difficulty with the above question so I googled and found

Atomic weights: -

Pb = 207,2
Sb = 121,76
O = 16

Let the mass of Sb be x grams
Then mass of Pb is (1,34 - x ) grams.

In Sb2O4 the ratio of O to Sb is the same as in SbO2

Therefore 121,76 g Sb combines with 32 g O2 and
207,2 grams of Pb combines with 32 grams of O2

So oxygen that reacts with Sb = 32x / 121,76
and O2 that combines with Pb = 32/207,2 X (1,34 - x)

Total combined O2 = 1,558 -1,340 = 0,218 g
Therefore: -
32x/121,76 + (1,34 -x) X 32/207,2 = 0,218
Simplify 32X207,2x + 5221,0688 - 3896,32x = 5499,85
6630,4x + 5221,0688 - 3896,32x = 5499,85

2734,08 x = 278,781
x = 0,10197 grams
Sb = 0,102 g
Pb = 1,238 g

Therefore Sb = 7,61%
I don't really understand this part:

So oxygen that reacts with Sb = 32x / 121,76
and O2 that combines with Pb = 32/207,2 X (1,34 - x)
Could someone please explain how to do this question?