Whatweight of oxygen will react with 1 g of calcium ? (Ca = 40)

1.	Whatweight of oxygen will react with 1 g of calcium ? (Ca = 40)
Answer» Solution :First Method Since all the atoms of Ca have CHANGED into CaO,the amount of Ca in CaO is 1 g. Nowfromthe 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 forCa atoms, molesof Ca in the reactant = moles of Ca in CaO `(1)/(40) ` (Rule 2)`= 1 xx ` moles of CaO (`:.` 1 mole of CoOcontains 1 mole of Ca atom ) `:. ` moles of CaO `= (1)/(40) "". . . (1) ` Againapplying POAC foroxygen atoms, Moles of O in `O_(2)` = moles of O inCaO `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 1moleof 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. Butin complicated reactions (Ex. s, 6, etc.) where the balancingis notvery easy , the studentcan applyPOAC withoutbalancingthe equation . Thisis where the mole methodhas its importance ]

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