The weight of oxygen that will react with 1 g of
calcium is

Explanation

First Method
Since all the atoms of Ca have changed into CaO, the amount of Ca in CaO is 1 g. Now from the formula of CaO , we have ,
moles of Ca = moles of O (Rule 6)
Now , `(wt. of Ca)/( at. wt. of Ca ) = (wt . of O)/( at. wt. of O) `
`:.` Wt. of oxygen `= (1)/(40) xx 16 = 0 . 4 `g .
second Method
`Ca + O_(2) to CaO` (balancing of the equation is not required)
1 g x g (say)
Applying POAC for Ca atoms,
moles of Ca in the reactant = moles of Ca in CaO
`(1)/(40) ` (Rule 2) `= 1 xx ` moles of CaO
(`:.` 1 mole of CoO contains 1 mole of Ca atom )
`:. ` moles of CaO `= (1)/(40) " ". . . (1) `
Again applying POAC for oxygen atoms,
Moles of O in `O_(2)` = moles of O in CaO
`2xx` moles of `O_(2) = 1 xx` moles of CaO . . . (2)
(`:.` 1 moles `O_(2)` contains 2 moles of O and 1 mole of CaO contains 1 mole of O).
From eqns (1) and (2) , eliminating moles of CaO, we have
moles of `O_(2) = (1)/( 2 xx 40 ) = (1)/(80) , " or " (wt. of O_(2))/( 32) = (1)/(80)" "` (Rule 1 )
`:.` wt . of `O_(2) = (1)/(8) xx 32 =0 . 4 g `
[Note : The chemical equation of the above given problem is simple, i.e., easy to balance. But in complicated reactions (Ex. s, 6, etc.) where the balancing is not very easy , the student can apply POAC without balancing the equation . This is where the mole method has its importance ]