(i) 1 gm of Mg is bumt in a closed vessel which contains `0.5` g of `O_(2)`.
(a) Which reactant is left excess.
(b) Find the mass of excess reactant.
(ii) A mixture of `KBr`, NaBr weighing `0.56` gm was treated with aquaeous solution of `Ag^(+)` and the bromide ion was recovered as `0.97` gm of pure AgBr. What was the weight of KBr in the sample ?

Explanation

(i) `{:(2Mg ,+,O_(2),to,2MgO),(2xx24,,2xx16,,2xx(24+16)),(=48 gm,,32gm,,=80 gm):}`
48 gm of Mg require oxygen ` = 32/48 = 0.66` gm
But only `0.5` g of oxygen is available . Hence `O_(2)` is limiting reagent and a part of magnesium wil not burn.
(b) 32 g of `O_(2)` react with magnesium 48 gm
` :. 0.5` gm `O_(2)` will react with magnesium ` = 48/32 xx 0.5 = 0.75` gm
Hence mass of magnesium left ` = (1 -0.75) = 0.25 `g
(ii) `{:(KBr,+,NaBr,+,"Ag"^(+),to,AgBr),(a gm ,,(0.56 -a)gm,,,,0.97 gm):}`
Applying POAC for Br atoms.
Moles of Br in KBr `+` Moles of Br in NaBr = Moles of Br AgBr
or ` 1 xx " Moles of " KBr + 1 xx " Moles of " NaBr = 1 xx " Moles of " AgBr`
`rArr a/119 + ((0.58 -a))/103 = (0.97)/188 (M_("KBr") = 199 , M_("NaBr") = 103 , M_("AgBr") = 183 )`
` :. a = 0.2125` gm
Percentage of KBr in the sample ` = (0.2125)/(0.56) xx 100 = 37 .9 %`