(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 ?

(i) 1 gm of Mg is bumt in a closed vessel which contains 0.5 g of O_(2

1.	(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 ?
Answer» Solution :(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 %`